My suggestion is to present 5^51 as 125^17, and 2^120 as 128^(120/7) which is 128^17.1.. . Since 128>125 and 17.1...>17 the second number is of course bigger. p.s your solution is very close, I solved it before watching)
Yea! This is kind of a rule, for two numbers say 'a' and 'b' more than 2, the rule applies where the lower value number with higher value power is always the greater than the higher value number with lower value power, where numbers are given in the form "a^b"and "b^a ". Note: a and b are to be same.i.e. either the base is to be same or power to be same for this rule to apply so in this case we gotta make power same or base same .
I solved using the fact that 2^10 is 1024, which is a little over 10^3 (it's a very important fact if you work with binary numbers a lot) we got 5^51 and 2^120, if we multiply both by 2^51 so we got 10^51 on one side [(2*5)^51] and 2^171 on the other since we know 2^10 > 10^3, then we can raise both sides to the power of 17, and we get 2^170 > 10^51 and we need to compare 10^51 to 2^171 2^171 > 2^170 > 10^51
It can be really simple by just taking log to the base e on both the sides... On lhs we'll have 51(ln 5) And rhs we'll have 120(ln2) We'll eventually get lhs as 82.11 and Rhs as 83.16. Clearly rhs>lhs Hence 2²¹⁰>5⁵¹...
I did it in my head. :) Tried the 5^50 first, saw that the inequality was going in the wrong direction. Then I realized 119 is a multiple of 17, so I went with 2^120 > 2^119 and this time the inequality came out right, (2^7)^17 > (5^3)^17
My attempt without watching/reading comments: is that I want a close inequality to try to bound this. 5^3 = 125 < 128 = 2^7, this is the smallest power that got a reasonably tight bound. Actually just for fun let's try a loose bound: 5^1 < 2^3, then we get 5^51 < 2^153, which is way too loose. Next 5^2 < 2^5, so 5*25^25 < 5*2^125 < 2^128, which is much better already. There is a clue built into the question since 51 is a power of 3 and 120 is a power of 8, but that could be misleading. Anyways, 125^17 < 2^(7*17) = 2^119 < 2^120. Edit: it would be difficult to get a tighter bound because 5^51 < 2^118.42
I did the same approach. It is easier to compare small numbers. And then you just need to increase the powers: 5^3 < 2^7 => (5^3)^17 < (2^7)^17 => 5^51 < 2^119 => 5^51 < 2^120 => DONE!
First thing I thought of was to take the cube root of both sides; this gives 5^17?2^40. Then factor 5^17 into (5^3)^5×5^2 and 2^40 into (2^7)^5×2^5. This gives 125^5×25
Very nice solution! You can also take the cube root of both to get 5^17 vs 2^40. Next notice that 2^10 = 1024, but just call it 1000 = 10^3. Now compare 5^17 vs 10^12. Divide both sides by 5^12 and compare 5^5 to 2^12. 3125 < 4096 so 2^120 is bigger.
If you can remember that log10(5) is 0.69897 and log10(2) is 0.30103 you are almost there. Using 0.7 and 0.3 as the values gives the log10 of the answers as: 357 and 360. As I rounded up the former and rounded down the latter that must indicate the answer.
Rewrite 5 as 10/2 and then take log base 10 on both sides and you will obtain 51(log10-log(2))= 51(1-log(2)), assuming we don’t know the value of log(2) we go back to the definition of the logarithm. Log(2) means what power must base 10 be raised to obtain 2 or another words 10^y=2. Note that 3/10
Awesome, Syber! Next up is e^pi vs pi^e, right? I plan to do a video on that sometime soon myself, so it will be interesting to see which approach we take :)
I did it this way: If (5 ^ 51) / (2 ^ 120) < 1 or > 1, then we know which is greater. = ((5/4) ^ 51 ) * (1 / (2^18)) = ((1.25) ^ 50) * (1.25) / (2 ^ 18) = ((1.5625) ^ 25) * (1.25) / (2 ^ 18) = (1.5625/2)^18 * (1.25^8) = ((0.78125)^18) * ((1.25)^8) = ((0.78125 * 1.25) ^ 8) * ((0.78125)^10) = ((0.9765625)^8) * ((0.78125)^10) As both the terms are less than 1, it means that product of their exponentials is also less than 1. QED
Alternate method: Take log on both sides. Compute its value by taking log 2~0.3010 and log 5~0.6989. Find the the number with larger value and you are done. log5⁵¹=35.6439 & log2¹²⁰=36.120
That would be true, but most people don't know what the value of log(2) and log(5) is, so we have to leave that as an exact answer since no calculators are allowed. However, we can substitute both of the log values for something else. We know that log(2)+log(5)=log(10)=1. Therefore, we can let one of the log values by comparing the other value. I'll try log(2)=1-log(5). We can tell that log(2)51log(5). Therefore, 5^51
@@albertmcchan I like your idea, but we have to remember that both log(2) and log(5) have an approximate answer, but since calculators aren't allowed and can't use in our own heads, we can just leave those as an exact answer. What you can tell from there is that 10 is the LCM of both 2 and 5 in order to make everything the same with the comparison of both values to make things a lot of easier since log(2)+log(5)=log(10)=1. That's why we have to substitute in one of those values in 120log(2) and 51log(5), and then go from there.
I took the cube root of both sides to produce 5^17 and 2^40. These were "smaller" numbers so it made comparing a little easier. 3,125^3 x 25 vs 1,024^4. 1,042^3 x 675 vs 1,024^4. If you take log base 2 of each, you get (estimates) 30.3 (generous) + 9.3 (guestimate) vs 40 (exact)
Congrats on reaching 25K subs. Again, a very nice video. For those of you young enough to consider math competitions, you may need to make this type of argument. (I only did one of these, a state competition, back in 1970!)
your first method works fine. 10th root of 5 is pretty small, close to 1, you could assume multiplying it only increases by 10%, while 2^12 is still ~33% larger than 5^5
i did binomial distribution , 5^51 can be written as (4+1)^51 and in case of 2^120 , 4^60, since there is 1 is case of (4+1) it becomes easier and hence we can clearly see which one is bigger , even if the nature of power(exponents) if you see in graph the curve is high so obv 2^120 is higher if u guessing
@@davidseed2939 some basic logarithm values are really convenient for estimating, but not many people know this lol I only memorized logarithm values for 2, 3, 7 though, but they're quite enough for most cases
Multiply by 2^51 on both sides. The reason for this is to convert 5^51 to a number of order 10. It becomes much easier to deal with. We are now comparing: 10^51 vs 2^171. Here’s the thing, I know that 2^10 is 1024, thus it is greater than 10^3. What’s so significant about this? Well, I know that (10^3)^17 = 10^51 but I also know that (2^10)^17 = 2^170 which is less than 2^171. But since 2^10 > 10^3 (as we’ve established) that must mean that 2^170 > 10^51. Thus meaning that 2^171 > 10^51 as well. As such 2^120 > 5^51, which is the final answer.
*Strategy* Get base close enough to each other by factoring the exponents - Since 51 = 3x17 and 5^3=125, therefore LHS = 5^51 = 125^17. Since 2^7 = 128 is close but greater than 125, therefore RHS = 2^120 = 128^(120/7) = 128^(17+1/7) Since both base (125 < 128) and exponent (17 < 17+1/7) are smaller, therefore LHS < RHS, hence *5^51 < 2^120*
The principle used to solve this problem is essentially the same as the one used to solve the problem "Comparing 31^11 and 17^14" in another video of yours. Both are based on showing the upper bound of one number is less than the exact value or the lower bound of another. The difference is how to do the bounding and this is greatly facilitated by the "nice relationship" between the numbers in the problems. For this video, it would be interesting to see how one can solve the problem without using a calculator if the second number is 2^118 instead of 2^120
You can solve using log_2 pretty easily. log_2(5^51) = 51log_2(4*5/4) = 102 + 51log_2(5/4). The rhs becomes simply 120. Now we have to figure out if 102 + 51log_2(5/4) < 120 => log_2(5/4) < 18/51. This is trhe since 18/51 > 1/3 > log_2(5/4). It is easy to see that log_2(5/4) < 1/3 since (5/4)^3 = 125/64 < 2.
I had a different approach for this one, I took 8 as a common factor and compared 2^14^8 with 5^6^8, 2^14=16384>15625=5^6 so 5^48 < 2^112, and we have that 2^8=256>125=5^3, so 5^51 < 2^120
Thanks for another awesome video! Personally the first thing that popped into my mind was to rewrite 5 as 2^(log of 5 base-2) so we get (2^(log5))^51 and it turns into a comparison of 51*log(5) and 120. But I'm seeing a lot of different methods, and I think some are better, and definitely don't require a calculator...
I went for 5^51 = 5*(5^5)^10 = 5*3125*10 vs (2^12)10 = 4096^10. Taking the tenth root on both sides and dividing by 3125, you get 4096/3125 = 1.310 vs 10rt(5). Finally, you can raise both sides to the 10th power and check if 1.31^10 is bigger than 5, which it is (in fact, 1.31^6 is already bigger) so 2^120 is bigger
My method would be to rewrite this as comparing 51*log base 2 of 5 with 120. Notice that 125 < 128, meaning that log base 2 of 5 < 7/3, so the left hand side is less than 119, implying that 5^51 < 2^119. Since 2^119 < 2^120, 5^51 < 2^120
My way is very heuristic. Take logs, order is preserved. We have 51ln5 v 120ln2 17ln5 v 40ln2=20ln4 ln 4 ~ ln5 (very small difference too lazy to evaluate through series) Meaning rhs is greater
I managed to do this one on my own: 5^51 vs 2^120 5^3=125 2^7=128 5^51 vs 2^120 (5^3)^17 vs (2^7)^17 x2 (125)^17 vs 2(128)^17 128>125 128^17>125^17 2(128)^17>125^17 Therefore 2^120 is greater. This is quite fun.
Nice solution from you thank you and it’s very interesting that in can be solved in a number of ways (some of which have been pointed out by other viewers)
Take log of both sides. Then use the power rule to get 51log(5) =? 120log(2). Rewrite to get log(5)/log(2) =? 120/51. Anyone with base 10 log experience knows that log(5) =.7 (.699 rounded off) and log(2) =.3 (.303 rounded off). .7/.3 = 7/3 = 2.33 and 120/51 is about 2.35. So, 5^51
Let's not think about what the approximation of log(2) and log(5) just because you can think in your head. Just leave those as an exact answer. What can you tell about log(2) and log(5) if there is something in common? The answer has to do with the LCM, which is 10. That means log(2)+log(5)=log(10)=1. Rearrange this and make that as a substitution as either 1-log(2)=log(5) or 1-log(2)=log(2) and then go from there.
@@robertveith6383 exactly. Although, I would not try guessing what the approximate answer of log(2) and log(5) is. Just go with the one where 2 and 5 have the LCM, which is 10. So, subsitute log(2)+log(5)=log(10)=1 as either 1-log(5)=log(2) or 1-log(2)=log(5) and then go from there.
For the summation 2+3^3+6+7^7+10+11^11+.............+(4x+2)+(4x+3)^(4x+3)=z^n, where n>1, how many +ve integer solutions for z are possible? Can you solve this
Taking log, and observing log function that it changes very low for a change in input, i.e log 2~~log 5, which makes 120 >>51, therefore 2^120 is greatest
But log 5 > log 4 = 2 log 2; your approximation is a bit too loose. In base 10, log 2 is close to 0.3 and log 5 is close to 0.7. 51 log 5 is about 35.7 while 120 log 2 is about 36. As others have pointed out (I've only looked at comments, haven't yet watched the video), 5^3 < 2^7, so 5^51 = (5^3)^17 < (2^7)^17 = 2^119 < 2^120.
@@tomkerruish2982 let's not carry out what the approximation of log(2), log(3), log(5), etc. just because you can think in your head, but we have to remember the log of something that is not powers of 10 is not an exact answer but an approximate answer. We can only leave that as log(2) and log(5) as an exact answer. However, both log(2) and log(5) have something in common, which is to take the LCM of both 2 and 5. That's 10. Using what we know from the properties of logarithms, log(2)+log(5)=log(10)=1. We can substitute one of the values in to make the comparison of another value to see which is bigger. You can say that 120log(2)=120(1-log(5))=120-120log(5). From there, we know that 120log(5)>51log(5). When you multiply a negative number, the answer is a negative number. Then, adding 120 to -120log(5) yields a positive number, so 120log(2)>51log(5). Therefore, this yields 2^120>5^51.
Try it with 5^59 vs. 2^137. 59*0.7 = 41.3. 137*0.3 = 41.1. That would imply that 5^59 > 2^137 but it is not. Your log approximations have enough error to not work when the 2 expressions are almost equal. This is cuz Log(2) ≈ .301 and Log(5) ≈ .699. 59*0.699 = 41.241 and 137*0.301 = 41.237 so even going to a 3rd decimal digit is not enough. 59*log(5) = 41.239... and 137*log(2) = 41.241... so now we can see 2^137 is larger than 5^59, but just barely. Also, in your solution, I think you mean 36 > 35.7.
We have to remember that log(2) and log(5) are irrational numbers. Therefore, you can only leave those as an exact answer. However, you can see that both log(2) and log(5) have something in common by taking the LCM of those values that is inside the log of an argument, which is 10. log(2)+log(5)=log(10)=1. We can substitute either 1-log(5)=log(2) or 1-log(2)=log(5). I'll go with the first choice: 120log(2)=120-120log(5), and then go from there.
@@justabunga1 My main point is that log approximations do not always work for problems like this, when both terms are very close, as they are in my example. I used 5^59 and 2^137 purposely to illustrate this. The difference between those is less than half of 1%. Also, I disagree with having to treat log(2) and log(5) as an exact answer only. In many (just not all) cases, the approximations of log(2) = 0.3 and log(5) = 0.7 are useful, especially when comparing 2 expressions that are fairly far apart in value (such as 5^51 and 2^120). My example is a counterexample where these approximations fail (lead you to believe that 5^59 is larger than 2^137, when actually 2^137 is slightly larger).
@@davidjames1684 we have to assume that log(2) and log(5) are unknown values to the approximation since both of them are not powers of 10, but I agree what you said so far. It's not like log(10)=1, log(100)=2, etc. and is between 0 and 1. Examples like pi, sqrt(2), or any irrational numbers are impossible to think this in your head without using your calculator. Anything that is an irrational number will always have to be left as an exact answer even though you can think this in your head as an approximate answer and can sometimes lead to different answers with the rounding and error, which is why I wouldn't do that. Something like larger numbers will have to use some other method that can be broken into some parts. Like you said 5^59 and 2^137 is impossible to think this in your head without the use of calculator, but we know is by the statement of equalities/inequalities: If log(a)>log(b), then a>b (in the same way that log(a)
The method used is universal. For your task, it was enough to use a logarithm with one decimal. For more accurate calculations, log (2) = 0.30103 ... log (5) = 1-0.30103 = 0.69897 log (2) is easy to remember. Draw one head. The left ear is 3, the left eye is 0, the nose is 1, the right eye is 0 and the right ear is 3. In my time of study, more than half a century ago, there were no computers. The memorized log (2) was useful to me in both chemical kinetics and radiochemistry.
I really enjoy ur way of thinking… and btw, I just use intuitive guess to get right this time… because this is why people want more “power” than a bigger but not big enough “base”. ;-)
2^120 is greater Answer 2^120 (6:59) 2^120 vs 5^51 5^51 can be written as 5^3^17 or 125^17 (3x17=51) o 2^120 can be written as 2^7^17.14 (since 7 x 17.14=120), but 2^7= 128, so 2^120= 128^17.14 128^17.14 is greater than 125^ 17 Therefore 2^120 is greater since it is equivalent to 128^17.14 PS thought I did it differently from you but realize that not that much as it was the same approach such as looking for a 2^n and 5^m which is close (125 and 128). The only difference was that you reduced 120 to 119 since both 57 and 119 are divisible by 17, and hoped to prove that if 5^51 (or 125^17) was less than 2^119 then it was less than 2^120. I, instead, used 120 and divided it by 7 to get 17.14 which is greater than 17 (yet, luckily, very very close to 17).
I looked at their digit count using formula Floor(log_10(n)) + 1 and found out 5^51 has 36 digits, 2^120 has 37 digits. So 2^120 > 5^51, although their count are almost equal so this was a close call
Great video as usual... could you make a video on Rolle's theorem, Lagrange's mean value theorem, and it's application in solving equations and such? I know what the statement says, but I'm unable to apply it, so any help is much appreciated. Thanks in advance :D
Another way to do it was to take the natural log of 5^51 and 2^120 and then compared the quantities. 51•ln(5)= 82.1 and 120•ln(2)= 83.2 and clearly 2^120 is greater. But I understand your proof was for people who don't have a mathematical background or studied it a long time ago.
First of all since in R+, x^4>y^4 x>y we just need to compare 5^13 and 2^30. 2^30/5^13=2^43/10^13 So the problem is equivalent to: is 2^43 bigger than 10^13? 2^43=(2^10)^4x2^3=(1024)^4x8. We already know that 2^43>1000^4x8=8x10^12 but this is not enough. We need to estimate 1024^4 more precisely. The problem can be rewritten: is 1024^4 bigger than 1.25x10^12? We are not going to calculate in detail: 1024^4
all i did was to make 5^51 as 5(5^50) and then do like the beginning with 5(3125^10) < 4096^10. by observation, 4096^10 will be significantly greater than 3125^10 so multiplying 5 to 3125^10 will not make it greater than 4096^10 Therefore, 5^51 is less than 2^120
thats nothing ... fellows you need to tetrate ! try tetration, aka hyper-power en.wikipedia.org/wiki/Tetration which is bigger? 2^^65536 or 10^^65533 ? ^^ means tetration and 2^^65536 is power-tower containing 65536 twos while 10^^65533 is power-tower containing 65533 of tens ... BUT there is top-associative (or whitten in row we can say left-associative) arrangement by definition (because only this kind of arrangement gives us the biggest value), example: ((2^3)^4)^5 is only 1152921504606846976 ( 19 digits), but 2^(3^(4^5)) has cca 10^488 digits !!!! so what is 2^^65536? it is insane left-associative power-tower 2^(2^(2^(2^(2^ .... ^2)))) with 65536 twos and 10^^65533 is insane left-associative power-tower 10^(10^(10^(10^(10^ .... ^10)))) with 65533 tens these numbers makes any brain to explode, it is absolutely useless to talk about number of digits that represents number of digits that represents number of digits of that numbers ... INSANE (but talking about BIG numbers thats only beginning) I can tell you 2^^65536 IS BIGGER than 10^^65533, yes shocking ... all of the twos are tens, but if there is slightly less tens (65533 vs. 65536) in the power-tower , the insane number is smaller
Again, take the log both sides for these two comparison as 51log(5) and 118log(2). Use the same method as before, and you will see that 118log(2) is bigger.
@@SingaporeSkaterSam think about it. Do you see anything in common with log(2) and log(5)? The answer has to do LCM, so the LCM of 2 and 5 is 10. That means log(2)+log(5)=log(10)=1. We can substitute one of the logs. I'll try 118log(2) as 118(1-log(5))=118-118log(5). Can you see the difference compared with 51log(5)? Here, we know that 118log(2)>51log(5). Therefore, 2^118>5^51.
EDIT: Man.. I must be really weird, because I don't think anyone else did it like this, lol. My solution was a little more work, but I think it shows it nicely: 5^51 = (5^6)^8*5^3 2^120 = (2^14)^8*2^8 5^6 = 15,625 < 2^14 = 16,384 5^6 < 2^14 => (5^6)^8 < (2^14)^8 5^3 = 125 < 2^8 = 256 (5^6)^8 < (2^14)^8 && 5^3 < 2^8 => (5^6)^8*5^3 < (2^14)^8*2^8 Thus, 5^51 < 2^120. It's probably (definitely) not a syntactically good proof, but it works.
If both numbers are bigger than e then the number with base closer to e is bigger e is approximately 2.718 so 2 is obviously close hence 2^120 is bigger this is solved with differentiations on a function I think I’m not sure tho but a youtuber called momeme made a video about it
My suggestion is to present 5^51 as 125^17, and 2^120 as 128^(120/7) which is 128^17.1.. . Since 128>125 and 17.1...>17 the second number is of course bigger. p.s your solution is very close, I solved it before watching)
me too
Me too
Likewise. Solved it in my head. Thougg I screwed up a little. I thought 128 was 2^6. Oops.
Yep. Almost could do this in my head this way; but not quite.
Nice!
Every time I see this kinda questions I always pick the higher exponent, it happens to work too often lol
Haha! That's cool!
It has to be sufficiently large enough
Yea! This is kind of a rule, for two numbers say 'a' and 'b' more than 2, the rule applies where the lower value number with higher value power is always the greater than the higher value number with lower value power, where numbers are given in the form "a^b"and "b^a ".
Note: a and b are to be same.i.e. either the base is to be same or power to be same for this rule to apply so in this case we gotta make power same or base same .
I solved using the fact that 2^10 is 1024, which is a little over 10^3 (it's a very important fact if you work with binary numbers a lot)
we got 5^51 and 2^120, if we multiply both by 2^51
so we got 10^51 on one side [(2*5)^51] and 2^171 on the other
since we know 2^10 > 10^3, then we can raise both sides to the power of 17, and we get
2^170 > 10^51
and we need to compare 10^51 to 2^171
2^171 > 2^170 > 10^51
That was the first thing I thought of as well.
I multiplied both sides with 2^51. And applied compaction (10^3)^17 vs 2*(2^10)^17
Do the same way😎
This is the way I solved it too
This one is definitely smarter and quicker
It can be really simple by just taking log to the base e on both the sides...
On lhs we'll have
51(ln 5)
And rhs we'll have
120(ln2)
We'll eventually get lhs as
82.11 and
Rhs as 83.16.
Clearly rhs>lhs
Hence 2²¹⁰>5⁵¹...
I think the fact that he went to this lenght was to try and avoid using a calculator
@@alimuhammadnasir1571 Oh yes.. it can be possible...👍
@@harshalkashyap8367 Bhai Bhai...🔥😂😂
Not true
I did it in my head. :) Tried the 5^50 first, saw that the inequality was going in the wrong direction. Then I realized 119 is a multiple of 17, so I went with 2^120 > 2^119 and this time the inequality came out right, (2^7)^17 > (5^3)^17
My attempt without watching/reading comments: is that I want a close inequality to try to bound this. 5^3 = 125 < 128 = 2^7, this is the smallest power that got a reasonably tight bound. Actually just for fun let's try a loose bound: 5^1 < 2^3, then we get 5^51 < 2^153, which is way too loose. Next 5^2 < 2^5, so 5*25^25 < 5*2^125 < 2^128, which is much better already.
There is a clue built into the question since 51 is a power of 3 and 120 is a power of 8, but that could be misleading. Anyways, 125^17 < 2^(7*17) = 2^119 < 2^120.
Edit: it would be difficult to get a tighter bound because 5^51 < 2^118.42
I did the same approach. It is easier to compare small numbers. And then you just need to increase the powers: 5^3 < 2^7 => (5^3)^17 < (2^7)^17 => 5^51 < 2^119 => 5^51 < 2^120 => DONE!
5³
First thing I thought of was to take the cube root of both sides; this gives 5^17?2^40. Then factor 5^17 into (5^3)^5×5^2 and 2^40 into (2^7)^5×2^5. This gives 125^5×25
Very nice solution! You can also take the cube root of both to get 5^17 vs 2^40. Next notice that 2^10 = 1024, but just call it 1000 = 10^3. Now compare 5^17 vs 10^12. Divide both sides by 5^12 and compare 5^5 to 2^12. 3125 < 4096 so 2^120 is bigger.
Nice!
If you can remember that log10(5) is 0.69897 and log10(2) is 0.30103 you are almost there.
Using 0.7 and 0.3 as the values gives the log10 of the answers as: 357 and 360.
As I rounded up the former and rounded down the latter that must indicate the answer.
Rewrite 5 as 10/2 and then take log base 10 on both sides and you will obtain 51(log10-log(2))= 51(1-log(2)), assuming we don’t know the value of log(2) we go back to the definition of the logarithm. Log(2) means what power must base 10 be raised to obtain 2 or another words 10^y=2. Note that 3/10
exactly :) I like your idea
Or without logs, note that 2^10 is just a bit bigger than 10^3, so writing 5=10/2 gives the ratio 2^120/5^51=2*(2^170/10^51)>2
Awesome, Syber! Next up is e^pi vs pi^e, right? I plan to do a video on that sometime soon myself, so it will be interesting to see which approach we take :)
Sounds good!
RPBP did this, a general rule of thumb for x^y>pi^e
I did it this way:
If (5 ^ 51) / (2 ^ 120) < 1 or > 1, then we know which is greater.
= ((5/4) ^ 51 ) * (1 / (2^18))
= ((1.25) ^ 50) * (1.25) / (2 ^ 18)
= ((1.5625) ^ 25) * (1.25) / (2 ^ 18)
= (1.5625/2)^18 * (1.25^8)
= ((0.78125)^18) * ((1.25)^8)
= ((0.78125 * 1.25) ^ 8) * ((0.78125)^10)
= ((0.9765625)^8) * ((0.78125)^10)
As both the terms are less than 1, it means that product of their exponentials is also less than 1. QED
Alternate method:
Take log on both sides. Compute its value by taking log 2~0.3010 and log 5~0.6989. Find the the number with larger value and you are done.
log5⁵¹=35.6439 & log2¹²⁰=36.120
That would be true, but most people don't know what the value of log(2) and log(5) is, so we have to leave that as an exact answer since no calculators are allowed. However, we can substitute both of the log values for something else. We know that log(2)+log(5)=log(10)=1. Therefore, we can let one of the log values by comparing the other value. I'll try log(2)=1-log(5). We can tell that log(2)51log(5). Therefore, 5^51
2^10 = 1024 ≈ 10^3, → log10(2) slightly above 0.3
log10(5) = log10(10/2), slightly below 1 - 0.3 = 0.7
log10(2^120) > 120 * 0.3 = 36.0
log10(5^51) < 51* 0.7 = 35.7 → 2^120 > 5^51
@@albertmcchan I like your idea, but we have to remember that both log(2) and log(5) have an approximate answer, but since calculators aren't allowed and can't use in our own heads, we can just leave those as an exact answer. What you can tell from there is that 10 is the LCM of both 2 and 5 in order to make everything the same with the comparison of both values to make things a lot of easier since log(2)+log(5)=log(10)=1. That's why we have to substitute in one of those values in 120log(2) and 51log(5), and then go from there.
I took the cube root of both sides to produce 5^17 and 2^40. These were "smaller" numbers so it made comparing a little easier.
3,125^3 x 25 vs 1,024^4.
1,042^3 x 675 vs 1,024^4.
If you take log base 2 of each, you get (estimates)
30.3 (generous) + 9.3 (guestimate) vs 40 (exact)
Congrats on reaching 25K subs. Again, a very nice video.
For those of you young enough to consider math competitions, you may need to make this type of argument. (I only did one of these, a state competition, back in 1970!)
your first method works fine. 10th root of 5 is pretty small, close to 1, you could assume multiplying it only increases by 10%, while 2^12 is still ~33% larger than 5^5
i did binomial distribution , 5^51 can be written as (4+1)^51 and in case of 2^120 , 4^60, since there is 1 is case of (4+1) it becomes easier and hence we can clearly see which one is bigger , even if the nature of power(exponents) if you see in graph the curve is high so obv 2^120 is higher if u guessing
51log5 = 51(0.6990) ~ 35
120log2 = 120(0.3010) > 36
Obviously, 2^120 is greater
i did it this way. quick and definite
@@davidseed2939 some basic logarithm values are really convenient for estimating, but not many people know this lol
I only memorized logarithm values for 2, 3, 7 though, but they're quite enough for most cases
Multiply by 2^51 on both sides. The reason for this is to convert 5^51 to a number of order 10. It becomes much easier to deal with.
We are now comparing:
10^51 vs 2^171.
Here’s the thing, I know that 2^10 is 1024, thus it is greater than 10^3.
What’s so significant about this? Well, I know that (10^3)^17 = 10^51 but I also know that (2^10)^17 = 2^170 which is less than 2^171. But since 2^10 > 10^3 (as we’ve established) that must mean that 2^170 > 10^51. Thus meaning that 2^171 > 10^51 as well.
As such 2^120 > 5^51, which is the final answer.
*Strategy* Get base close enough to each other by factoring the exponents -
Since 51 = 3x17 and 5^3=125, therefore LHS = 5^51 = 125^17.
Since 2^7 = 128 is close but greater than 125, therefore RHS = 2^120 = 128^(120/7) = 128^(17+1/7)
Since both base (125 < 128) and exponent (17 < 17+1/7) are smaller, therefore LHS < RHS, hence *5^51 < 2^120*
Find log base 2 of 5, multiply that by 51. compare that result with 120. Alternately, (ln 5) * 51 compared to (ln 2) * 120, whichever is larger.
The principle used to solve this problem is essentially the same as the one used to solve the problem "Comparing 31^11 and 17^14" in another video of yours. Both are based on showing the upper bound of one number is less than the exact value or the lower bound of another. The difference is how to do the bounding and this is greatly facilitated by the "nice relationship" between the numbers in the problems. For this video, it would be interesting to see how one can solve the problem without using a calculator if the second number is 2^118 instead of 2^120
5⁵¹ = (5³)¹⁷ = 125¹⁷
2¹²⁰ = 2 × 2¹¹⁹ = 2 × (2⁷)¹⁷ = 2 × 128¹⁷
Clearly the second is larger.
You can solve using log_2 pretty easily. log_2(5^51) = 51log_2(4*5/4) = 102 + 51log_2(5/4). The rhs becomes simply 120. Now we have to figure out if 102 + 51log_2(5/4) < 120 => log_2(5/4) < 18/51. This is trhe since 18/51 > 1/3 > log_2(5/4). It is easy to see that log_2(5/4) < 1/3 since (5/4)^3 = 125/64 < 2.
I had a different approach for this one, I took 8 as a common factor and compared 2^14^8 with 5^6^8, 2^14=16384>15625=5^6 so 5^48 < 2^112, and we have that 2^8=256>125=5^3, so 5^51 < 2^120
Sir,using logarithms ,we find that2^120 > 5^51as mantissa of 120x0.3010 > mantissa of 51x.6998(log5)
Thanks for another awesome video! Personally the first thing that popped into my mind was to rewrite 5 as 2^(log of 5 base-2) so we get (2^(log5))^51 and it turns into a comparison of 51*log(5) and 120. But I'm seeing a lot of different methods, and I think some are better, and definitely don't require a calculator...
I went for 5^51 = 5*(5^5)^10 = 5*3125*10 vs (2^12)10 = 4096^10. Taking the tenth root on both sides and dividing by 3125, you get 4096/3125 = 1.310 vs 10rt(5). Finally, you can raise both sides to the 10th power and check if 1.31^10 is bigger than 5, which it is (in fact, 1.31^6 is already bigger) so 2^120 is bigger
Very good explanation ! Thank you very much ! Video is useful for novice teachers .
Glad it was helpful!
My method would be to rewrite this as comparing 51*log base 2 of 5 with 120. Notice that 125 < 128, meaning that log base 2 of 5 < 7/3, so the left hand side is less than 119, implying that 5^51 < 2^119. Since 2^119 < 2^120, 5^51 < 2^120
Awesome problem !!! i like these types 😍
this is a joke btw.....
5^51-2^120 = -8.8513879 x 10^35 so clearly.....2^120 is bigger
Lol
*5^51 = 125^17 < 128^17 = 2^119 < 2^120, hence 5^51 < 2^120*
Your approach is the best , dear Vishal
My way is very heuristic. Take logs, order is preserved. We have
51ln5 v 120ln2
17ln5 v 40ln2=20ln4
ln 4 ~ ln5 (very small difference too lazy to evaluate through series)
Meaning rhs is greater
Find value of 51log5 and 128log2
Then the logarithm value of second number will be larger
So second number will be larger
Exactly xd
I think the point is to not use a calculator, unless u know log2 and log5 off the top of ur head
@@rishabhsetty3109 hmmm
Ok i Know the values of log2,3,5,7
It helps me
Yeah
This just means justice to integers.... Doesn't matter if it's in power or not
I liked your solution, it taught me alot! Thanks for the video!
Glad it helped!
6:34 my life changed all of a sudden 😂
😂
Interesting ~ so important to see the detail
5^51=125^17
Nice :S
:D
Simple and great!
I managed to do this one on my own:
5^51 vs 2^120
5^3=125
2^7=128
5^51 vs 2^120
(5^3)^17 vs (2^7)^17 x2
(125)^17 vs 2(128)^17
128>125
128^17>125^17
2(128)^17>125^17
Therefore 2^120 is greater. This is quite fun.
This video is Greater than both values lol.
Also Im glad adway kumar commented too.
Nice solution from you thank you and it’s very interesting that in can be solved in a number of ways (some of which have been pointed out by other viewers)
Take log of both sides. Then use the power rule to get 51log(5) =? 120log(2). Rewrite to get log(5)/log(2) =? 120/51. Anyone with base 10 log experience knows that log(5) =.7 (.699 rounded off) and log(2) =.3 (.303 rounded off). .7/.3 = 7/3 = 2.33 and 120/51 is about 2.35. So, 5^51
I memorized log(2) = 0.30 (approximately, of course). Then log(5) = 1-log(2) = 0.7
So 5 = 2^(0.7/0.3) = 5^(7/3)
5^51 = 2^(7*51/3) = 2^119
2^119
Let's not think about what the approximation of log(2) and log(5) just because you can think in your head. Just leave those as an exact answer. What can you tell about log(2) and log(5) if there is something in common? The answer has to do with the LCM, which is 10. That means log(2)+log(5)=log(10)=1. Rearrange this and make that as a substitution as either 1-log(2)=log(5) or 1-log(2)=log(2) and then go from there.
Using logatlritm:
5^51 ~ 2^120
Log (5^51) ~ Log(2^120)
51 log 5 ~ 120 log 2
51 log 5 / log 2 ~ 120
Since log 4/log 2 < log 5/log 2 < log 8/log 2, means that 2
No, you cannot guess/estimate 2.3 for log(5)/log(2). What if the guess is 2.36? Then 51*2.36 =
120.36 > 120. You are going to have to fix your proof.
@@robertveith6383 exactly. Although, I would not try guessing what the approximate answer of log(2) and log(5) is. Just go with the one where 2 and 5 have the LCM, which is 10. So, subsitute log(2)+log(5)=log(10)=1 as either 1-log(5)=log(2) or 1-log(2)=log(5) and then go from there.
We can divide both sides by 3125 ^10 so we get 1
compi outputs a True for this hehe:
*5^51 < 2^120*
I can see your statement, hehe.
I can't see how you did
@@keescanalfp5143 i never do. i have compi for this. 😁
2^120 is bigger
Take log of both expressions and sub value of log 2 and log 5 and you will see 120 log 2> 51 log 5 so 2^120 is bigger
For the summation 2+3^3+6+7^7+10+11^11+.............+(4x+2)+(4x+3)^(4x+3)=z^n, where n>1, how many +ve integer solutions for z are possible? Can you solve this
doesn't make sense. Maybe if you pull up lg 5 = 1 - lg2. And try to prove (120 + 51) lg2 > 51
Tough to do without a calculator. At least for me.
@@sujanshankarbhowmick9381 I think the answer is 0
I used this method too. log 2 is about 0.3 and log 5 is about 0.7. So 120 log 2 is about 36, and 51 log 5 is about 35.7. So 2^120 is larger.
Taking log, and observing log function that it changes very low for a change in input, i.e log 2~~log 5, which makes 120 >>51, therefore 2^120 is greatest
But log 5 > log 4 = 2 log 2; your approximation is a bit too loose. In base 10, log 2 is close to 0.3 and log 5 is close to 0.7. 51 log 5 is about 35.7 while 120 log 2 is about 36.
As others have pointed out (I've only looked at comments, haven't yet watched the video), 5^3 < 2^7, so 5^51 = (5^3)^17 < (2^7)^17 = 2^119 < 2^120.
@@tomkerruish2982 let's not carry out what the approximation of log(2), log(3), log(5), etc. just because you can think in your head, but we have to remember the log of something that is not powers of 10 is not an exact answer but an approximate answer. We can only leave that as log(2) and log(5) as an exact answer. However, both log(2) and log(5) have something in common, which is to take the LCM of both 2 and 5. That's 10. Using what we know from the properties of logarithms, log(2)+log(5)=log(10)=1. We can substitute one of the values in to make the comparison of another value to see which is bigger. You can say that 120log(2)=120(1-log(5))=120-120log(5). From there, we know that 120log(5)>51log(5). When you multiply a negative number, the answer is a negative number. Then, adding 120 to -120log(5) yields a positive number, so 120log(2)>51log(5). Therefore, this yields 2^120>5^51.
log(2) ≈ 0.3, then log(5)=0.7 120 log (2) vs. 51 log (5), 360 > 357 😍
Try it with 5^59 vs. 2^137. 59*0.7 = 41.3. 137*0.3 = 41.1. That would imply that 5^59 > 2^137 but it is not. Your log approximations have enough error to not work when the 2 expressions are almost equal. This is cuz Log(2) ≈ .301 and Log(5) ≈ .699. 59*0.699 = 41.241 and 137*0.301 = 41.237 so even going to a 3rd decimal digit is not enough. 59*log(5) = 41.239... and 137*log(2) = 41.241... so now we can see 2^137 is larger than 5^59, but just barely. Also, in your solution, I think you mean 36 > 35.7.
We have to remember that log(2) and log(5) are irrational numbers. Therefore, you can only leave those as an exact answer. However, you can see that both log(2) and log(5) have something in common by taking the LCM of those values that is inside the log of an argument, which is 10. log(2)+log(5)=log(10)=1. We can substitute either 1-log(5)=log(2) or 1-log(2)=log(5). I'll go with the first choice: 120log(2)=120-120log(5), and then go from there.
@@justabunga1 My main point is that log approximations do not always work for problems like this, when both terms are very close, as they are in my example. I used 5^59 and 2^137 purposely to illustrate this. The difference between those is less than half of 1%. Also, I disagree with having to treat log(2) and log(5) as an exact answer only. In many (just not all) cases, the approximations of log(2) = 0.3 and log(5) = 0.7 are useful, especially when comparing 2 expressions that are fairly far apart in value (such as 5^51 and 2^120). My example is a counterexample where these approximations fail (lead you to believe that 5^59 is larger than 2^137, when actually 2^137 is slightly larger).
@@davidjames1684 we have to assume that log(2) and log(5) are unknown values to the approximation since both of them are not powers of 10, but I agree what you said so far. It's not like log(10)=1, log(100)=2, etc. and is between 0 and 1. Examples like pi, sqrt(2), or any irrational numbers are impossible to think this in your head without using your calculator. Anything that is an irrational number will always have to be left as an exact answer even though you can think this in your head as an approximate answer and can sometimes lead to different answers with the rounding and error, which is why I wouldn't do that. Something like larger numbers will have to use some other method that can be broken into some parts. Like you said 5^59 and 2^137 is impossible to think this in your head without the use of calculator, but we know is by the statement of equalities/inequalities: If log(a)>log(b), then a>b (in the same way that log(a)
The method used is universal. For your task, it was enough to use a logarithm with one decimal.
For more accurate calculations, log (2) = 0.30103 ... log (5) = 1-0.30103 = 0.69897 log (2) is easy to remember. Draw one head. The left ear is 3, the left eye is 0, the nose is 1, the right eye is 0 and the right ear is 3. In my time of study, more than half a century ago, there were no computers. The memorized log (2) was useful to me in both chemical kinetics and radiochemistry.
It helpe us reckon the long looking calculation instead shortcut to answer
Thank you from a math teacher in indonesia
I really enjoy ur way of thinking… and btw, I just use intuitive guess to get right this time… because this is why people want more “power” than a bigger but not big enough “base”. ;-)
Or just change 2 to 4 and then log it
So it would be 60log4 and 51 log5
2^120 is greater Answer 2^120 (6:59)
2^120 vs 5^51
5^51 can be written as 5^3^17 or 125^17 (3x17=51) o
2^120 can be written as 2^7^17.14 (since 7 x 17.14=120), but 2^7= 128, so 2^120= 128^17.14
128^17.14 is greater than 125^ 17 Therefore 2^120 is greater since it is equivalent to 128^17.14
PS thought I did it differently from you but realize that not that much as it was the same approach such as looking for a 2^n and 5^m which is close (125 and 128). The only difference was that you reduced 120 to 119 since both 57 and 119 are divisible by 17, and hoped to prove that if
5^51 (or 125^17) was less than 2^119 then it was less than 2^120. I, instead, used 120 and divided it by 7
to get 17.14 which is greater than 17 (yet, luckily, very very close to 17).
I looked at their digit count using formula Floor(log_10(n)) + 1 and found out 5^51 has 36 digits, 2^120 has 37 digits. So 2^120 > 5^51, although their count are almost equal so this was a close call
Great video as usual... could you make a video on Rolle's theorem, Lagrange's mean value theorem, and it's application in solving equations and such? I know what the statement says, but I'm unable to apply it, so any help is much appreciated. Thanks in advance :D
Bump
That's math with common sense . Awesome job 👌
For the summation 2+3^3+6+7^7+10+11^11+.............+(4x+2)+(4x+3)^(4x+3)=z^n, where n>1, how many +ve integer solutions for z are possible?
O
なんとか粘って考え,
5^51 , 2^120 = √5^102 , √2^240
= (√5^3)^34 , (√2^7)^34.2857…
= √125^34 ＜ √128^34.2857…
となりました。
高校入試 数学 でもいけますね。
Well my solution is:
5^3 is less than 2^7
(5^3)^17 is less than (2^7)^17
5^51 is less than 2^119
5^51 is less than 2^120
Another way to do it was to take the natural log of 5^51 and 2^120 and then compared the quantities. 51•ln(5)= 82.1 and 120•ln(2)= 83.2 and clearly 2^120 is greater. But I understand your proof was for people who don't have a mathematical background or studied it a long time ago.
This eliminates the need for the memorization of the values of log2 and/or log5 or the use of calculators
My solution:
5^10 = 9765625 < 10^7
5^51 = 5*(5^10)^5
5^51 < 5*(10^7)^5
5^51 < 5*10^35
2^10 = 1024 > 10^3
2^120 = (2^10)^12
2^120 > (10^3)^12
2^120 > 10^36
All together, this means that 5^51 < 5*10^35 < 10^36 < 2^120, implying that 5^51 < 2^120.
notice that 2^7=128>125=5^3, hence 2^120>2^119=(2^7)^17>(5^3)^17=5^51
understand the steps perfectly bro, thanks for sharing
No problem 👍
Is using AM => GM inequality possible?
Multiply both sides by 2^51
You compare 10^51 and 2^171
2^10 = 1024 > 10^3
2^171 > (10^3)^17,1 = 10^51,3 > 10^51
Very very good
Thanks
Can I use logarithms?
Nice vid! Please do a comparison between 5^52 and 2^120
Thank you!
First of all since in R+, x^4>y^4 x>y we just need to compare 5^13 and 2^30.
2^30/5^13=2^43/10^13
So the problem is equivalent to: is 2^43 bigger than 10^13?
2^43=(2^10)^4x2^3=(1024)^4x8.
We already know that 2^43>1000^4x8=8x10^12 but this is not enough. We need to estimate 1024^4 more precisely.
The problem can be rewritten: is 1024^4 bigger than 1.25x10^12?
We are not going to calculate in detail: 1024^4
seriously finally this one....
I loved this method!
Glad you liked it!
Well, if I had a calculator I would evaluate their logarithms 😎
At the end of the day that is the reason logarithms exist!
Nice video!
Thank you!
Good question of this type, its incredible that's 2^120/5^51≈3
all i did was to make 5^51 as 5(5^50) and then do like the beginning with 5(3125^10) < 4096^10. by observation, 4096^10 will be significantly greater than 3125^10 so multiplying 5 to 3125^10 will not make it greater than 4096^10
Therefore, 5^51 is less than 2^120
I'd be highly surprised if a 120th power of a number were to be smaller than a 51th power of a slightly bigger number.
I solved like -> 5^51 = (4 x 1.25)^51 = 2^102 x (1.25)^51 = 2^102 x (1.25^3)^17 = 2^102 x (1.953..)^17 < 2^102 x 2 ^ 18 = 2^120
2^120>2^119
2^120)>128^17
5^51=125^17
So, 5+511×2+120>0^3=:"120 Zeros with a 2 after it is larger than 5 with 51 zeros after it.
Compare log to log:51*log(5) versus 2*log(120). Easy calculation by excel
1:38
Technically it *would* help if you could show that 5
2^120 is a easy, we also know is smaller than 2^119, but is it smaller than 2^118?
thats nothing ... fellows you need to tetrate ! try tetration, aka hyper-power en.wikipedia.org/wiki/Tetration
which is bigger? 2^^65536 or 10^^65533 ?
^^ means tetration and 2^^65536 is power-tower containing 65536 twos while 10^^65533 is power-tower containing 65533 of tens ... BUT there is top-associative (or whitten in row we can say left-associative) arrangement by definition (because only this kind of arrangement gives us the biggest value), example: ((2^3)^4)^5 is only 1152921504606846976 ( 19 digits), but 2^(3^(4^5)) has cca 10^488 digits !!!!
so what is 2^^65536? it is insane left-associative power-tower 2^(2^(2^(2^(2^ .... ^2)))) with 65536 twos
and 10^^65533 is insane left-associative power-tower 10^(10^(10^(10^(10^ .... ^10)))) with 65533 tens
these numbers makes any brain to explode, it is absolutely useless to talk about number of digits that represents number of digits that represents number of digits of that numbers ... INSANE (but talking about BIG numbers thats only beginning)
I can tell you 2^^65536 IS BIGGER than 10^^65533,
yes shocking ... all of the twos are tens, but if there is slightly less tens (65533 vs. 65536) in the power-tower , the insane number is smaller
Take nature log on both sides 51ln5
nice trick sir thanks
2pow2.32 is almost 5. So I can write 2pow(2.32*51) which will be almost 2pow118. So 2pow120 will be greater.
the task is more difficult 5^51 or 2^118 ?
Again, take the log both sides for these two comparison as 51log(5) and 118log(2). Use the same method as before, and you will see that 118log(2) is bigger.
@@justabunga1 run that by me again…
@@SingaporeSkaterSam think about it. Do you see anything in common with log(2) and log(5)? The answer has to do LCM, so the LCM of 2 and 5 is 10. That means log(2)+log(5)=log(10)=1. We can substitute one of the logs. I'll try 118log(2) as 118(1-log(5))=118-118log(5). Can you see the difference compared with 51log(5)? Here, we know that 118log(2)>51log(5). Therefore, 2^118>5^51.
5^1 and 2^2 are closer to each other than 5^3 and 2^7...
EDIT: Man.. I must be really weird, because I don't think anyone else did it like this, lol.
My solution was a little more work, but I think it shows it nicely:
5^51 = (5^6)^8*5^3
2^120 = (2^14)^8*2^8
5^6 = 15,625 < 2^14 = 16,384
5^6 < 2^14 => (5^6)^8 < (2^14)^8
5^3 = 125 < 2^8 = 256
(5^6)^8 < (2^14)^8 && 5^3 < 2^8 => (5^6)^8*5^3 < (2^14)^8*2^8
Thus, 5^51 < 2^120.
It's probably (definitely) not a syntactically good proof, but it works.
It's pretty good!
@@SyberMath Hey, thanks!
nice!
Bro I am from India and I am in grade 12 . This prob took me 15 sec .simply take log of both numbers
And you will get your answer
ok. Do you use a calculator?
For what purpose
@@lucifer8896 For the logs
No actually i am a JEE ADVANCED aspirant so we have to learn the basic values. 😀
Why not take log of both sides to base 2. Log(5) will be 2.32....
51*2.32... is less 120 and we are done.
take log of 2 with base 5 and you will see that it comes out as 2¹¹⁸ which is less than 2¹²⁰ so b is greater
Both the values lie in x if x^0 -0^x = 1 over and out 🙂
If both numbers are bigger than e then the number with base closer to e is bigger e is approximately 2.718 so 2 is obviously close hence 2^120 is bigger this is solved with differentiations on a function I think I’m not sure tho but a youtuber called momeme made a video about it
my suggestion is to compare 5(5^50) and 16^50
Why not just see that 5=2^(sqrt(5))? Then argue that 2
Because it's not true. 5 is not 2 to the power of root 5, which is transcendental.
Using logs,
2^120=5^(51.68) which is greater than 5^51
Why not use logs ?
Because we want to do it without using a calculator