It's the opposite of exponenciation, your x in a logarithm is the exponent you need to give to the base (10 in this case) so that the result is equal to 80 (which is called the argument I think). Not that hard to grasp as a concept
i did it by applying log to both sides, and then solving. it makes it easier because after that its just manipulation. but you do need the value for log 2 and log 40.
Before even watching the video I know I'm gonna hear "Lets put a box around it" and "How exciting". Almost forgot the “Nice” every time 69 or 420 shows up. And you can’t forget the “this looks like a fun problem” when the video starts.
For the people trying to show off your logarithmic skills, Andy intended this problem to be understood by those who aren't really familiar with log. So, instead of using log identities from the get-go, he used log only when the equation was already simplified. You and I may be familiar with log, not everyone is.
You could also solve this problem using logarithmic properties: 1. We take the log() of both sides, which allows us to put the exponent as a factor to the logarithm of the base. Giving us: "(x-1)log(40)=(2x+1)log(2)" 2. We multiply into the parenthesis on both sides, and isolate the terms with x in them. Giving us: "x log(40)-2x log(2)=log(2)+log(40)" 3. We isolate x by factoring, and dividing the resulting parenthesis to the right. Giving us: "x = (log(2)+log(40)) / (log(40)-2log(2)) 4. Due to the rule of logarithms used in (1), we can express the "2log(2)" as "log(2^2)" which equals "log(4)", whereafter we use the rule of sums and differences of logarithms. Giving us: "x=log(80)/log(10)" 5. Due to log-base10 of 10 being 1, this gives us the result as: "x=log(80)"
My math teacher in high school wouldn't have allowed log 80 as an answer because it can be reduced like square root of 80 can be reduced. Because it is base 10, log 80 = log 10 + log 8 since log ab = log a + log b if the base is the same. Log 10 = 1, so log 80 = 1+ log 8.
Depends, sometimes it’s the opposite. With a class that I was in, they would want me to write 1 + log 8 as log 80. But log 8 on its own would need to be simplified to 3log 2.
I tried solving it before the video, ended up with the same answer. I just did the log base 2 of 40 first to get the exponent bases to 2. got something around 5.32. set the exponents = to each other due to same bases and boom basic algebra to get to the same decimal answer. (with calculator ofc) nice video tho man 👍
I’m having flashbacks to helping my daughter with Algebra 2 😮 I love how you manage to keep the presentation so brisk, yet also don’t have any mystery steps.
A video about a list of the usual "tricks" that can be used in such math problems (like the 30-60-90 triangle, the exponent substitutions you used here, etc...) would be amazing! Thanks!
Easy method here Just take log on both sides and multiply it in side then you should have an equation with both sides having x of power one and substitute values of log40 ( log4 +log10) and log 2 Then u get 1.602x-1.602=0.601x+0.301 On bringing x oneside and constant on other side u get 1.001x=1.903 Which is approx answer x=1.903 If you need more precise then increase the decimal numbers That's it not even a 2min task
That's what i was thinking it took me about 1 min to solve this mentally i was like he must have asked a numerical value i solved log 8 ans 3log2 = 3X.30 and i got 1.90
This is the first time I watch your videos and honestly they're actually helping me understand math even better, I don't know, I try to resolve the problem a little faster than you just to see if I can do it, sometimes I get it wrong but it's fine, I keep trying, thank you! you're helping someone get better at math!
2:31 I think explaining that with the inverse function is more logical. Just like in _tan x = 2_ we take the inverse function of _tan_ on both sides, which is _tan^-1_ that gives us _tan^-1 (tan x) = tan^-1 (2)_ which simplifies to _x = tan^-1 (2)_ Here, the inverse function of _10^x_ is _log 10 (x)_ Taking that on both sides gives us _log 10 (10^x) = log 10 (80)_ which simplifies to _x = log 10 (80)_
@@rathersane You can use logarithm regardless but yes, it's usually 10 as base. In this case, it works better. I'm just saying what I prefer in my opinion. Is that bad?
Hi man! What an amazing video! But, the verification can be done like this without a calculator: 40^(x - 1) = 40^(log80 - 1) = 40^(log8 + log10 - 1) = 40^(log8 + 1 - 1) = 40^log8 2^(2x + 1) = 2^(2log80 + 1) = 2^(2(log8 + log10) + 1) = 2^(2(log8 + 1) + 1) = 2^(2log8 + 2 + 1) = 2^(2log8 + 3) = 2^2log8 * 2^3 = 4^log8 * 8 = 4^log8 * 10^log8 = (4*10)^log8 = 40^(log8) here, we can see that 40^(x - 1) gave us 40^log8 and 2^(2x + 1) also gave us 40^log8. so, verified.
I knew that because I brute forced and put 1 as x and 2 as x and then just did 1.5 and kept raising it to 1.9.... They need to make it a bigger x so that I can't just input numbers quicker than it takes to actually use the correct formulas😅
just introduce logs to both sides in the beginning 40^x-1=2^2x+1 log 40^x-1= log 2^2x+1 power law of logs, the power of the log can become the coefficient of the log (x-1) log 40 = (2x+1) log 2 x log40 - log40 = 2x log2 + log2 move the x terms to the left, and the log40 to the right x log40 - 2x log2 = log 2 + log 40 factorise x from the right x(log 40 -2log2) = log2 + log40 x= (log2 + log40) / (log40 -2log2) = 1.903089987
There was a really easier way, you could write as (2^ 5.32192809490)^x-1=2^2x+1 ,since both bases are the same we can write down as 5.3219280949(x-1)=2x+1 which is then equal to 5.3219280949x-2x= 5.3219280949+1 then x= 6.3219280949/3.3219280949
so simple you just know about log 5on2 =2.3 and done and solve it but you can do it simple you should change this log to log5 /log2 => 0.7/0.3 =2.3 (rounded ) and you dont wana calculator or somthing this question was standard for student
From 1st sec, i striaght knows that just using ln() then find the x. Just that simple. No need to be simplify🤔 why need sinplify as we can just using ln() loll
problem solving is a very important transferable skill. Being good at solving any math problem means your likely a good problem solver in any setting. Yes, you won't need to solve this specific equation or anything like this in a practical setting, but it trains your brain to think outside the box
It’s an exponential function. You can google the purpose and application of those. This is a problem solving channel, not an application and vague question answering channel.
by trial and error i found 1 then 2 then 1.9. and 1.9 is way way too close. I found in 3rd iteration via calculator. I will take that as ok becuz in real life that would negligible difference, mostly covered by buffer and in real life u won't be fucking sitting there finding value of log 69 or something. Maths doesn't has real life applications beyond class 10 or 12 . Edit: the answer is 1.90309 lol That's 0.16 % difference.
![[UNCUT] "เสรีพิศุทธ์" แฉยับ! งัดหลักฐานใหม่ชั้น 14 ได้นอนคุกสมใจแน่!! I คนดังนั่งเคลียร์ I 2 ก.ย.67](http://i.ytimg.com/vi/O1SSg4p9Izs/mqdefault.jpg)
3:37 i think you meant to say you rounded not truncated :)
Oh no! You are right! I messed that up...
I see, that's why I'm a bit confused 😂
@AndyMath all good, love the videos!
Truncated means rounded off. Both are same dear
@@insaneclips1184 no, truncated means you cut off additional digits (usually decimal). So 76.73 would truncate to 76 but round to 77.
everyone is gangsta untill log shows up
True af
Perfect comment😂😂
Kids!!
bro lowkey
log is taught in 9th grade though. it's just BAP.
My mind after 0.00000001 seconds of looking at this equation : Log Log Log Log
fr same
idek what log is i'm doing alg 1 rn
Fr
It's the opposite of exponenciation, your x in a logarithm is the exponent you need to give to the base (10 in this case) so that the result is equal to 80 (which is called the argument I think). Not that hard to grasp as a concept
Exactly! Log from line 1
India's secondary school students be like,"like, seriously‽."
Really!!
Yep, its kinda easy
Mein timepass karne TH-cam mein aaya aur yaha aake yeh problem solve karne baith gya 😂
Even though it is easy for us it may not be the case for whole India. So we should not be full of ourselves.
So true
Idk why, but this felt like pulling off a combo KO in YuGiOh
Bro why ain't I ever thought bout learning math as learning combos in fighting games dawg
@@Redawesomeoby Hadouken to find x 😂😂😂
i did it by applying log to both sides, and then solving. it makes it easier because after that its just manipulation. but you do need the value for log 2 and log 40.
Need to know just log 2.
log 40 would be log 4 + log 10 . Everyone knows log 10 value and log 4 would be 2 × log 2.
Before even watching the video I know I'm gonna hear "Lets put a box around it" and "How exciting".
Almost forgot the “Nice” every time 69 or 420 shows up. And you can’t forget the “this looks like a fun problem” when the video starts.
how exciting
This looks like a fun comment.
it’s pure hype
And it looks important, so let's put a box around it 😅
Or one can just use log
For the people trying to show off your logarithmic skills, Andy intended this problem to be understood by those who aren't really familiar with log. So, instead of using log identities from the get-go, he used log only when the equation was already simplified. You and I may be familiar with log, not everyone is.
Wonderful explanation.This man can teach us faster than a professional professor.
I think he is a professional
He is a teacher
A professor is inherently a professional
Definitely a middle school teacher @@euloge996
You could also solve this problem using logarithmic properties:
1. We take the log() of both sides, which allows us to put the exponent as a factor to the logarithm of the base. Giving us: "(x-1)log(40)=(2x+1)log(2)"
2. We multiply into the parenthesis on both sides, and isolate the terms with x in them. Giving us: "x log(40)-2x log(2)=log(2)+log(40)"
3. We isolate x by factoring, and dividing the resulting parenthesis to the right. Giving us: "x = (log(2)+log(40)) / (log(40)-2log(2))
4. Due to the rule of logarithms used in (1), we can express the "2log(2)" as "log(2^2)" which equals "log(4)", whereafter we use the rule of sums and differences of logarithms. Giving us: "x=log(80)/log(10)"
5. Due to log-base10 of 10 being 1, this gives us the result as: "x=log(80)"
My math teacher in high school wouldn't have allowed log 80 as an answer because it can be reduced like square root of 80 can be reduced. Because it is base 10, log 80 = log 10 + log 8 since log ab = log a + log b if the base is the same. Log 10 = 1, so log 80 = 1+ log 8.
1+3log2 which is approx 1.9
Depends, sometimes it’s the opposite. With a class that I was in, they would want me to write 1 + log 8 as log 80. But log 8 on its own would need to be simplified to 3log 2.
I tried solving it before the video, ended up with the same answer. I just did the log base 2 of 40 first to get the exponent bases to 2. got something around 5.32. set the exponents = to each other due to same bases and boom basic algebra to get to the same decimal answer. (with calculator ofc) nice video tho man 👍
I’m having flashbacks to helping my daughter with Algebra 2 😮
I love how you manage to keep the presentation so brisk, yet also don’t have any mystery steps.
An excellent demonstration ! That was clear and easy to follow. Thanks for sharing !
A video about a list of the usual "tricks" that can be used in such math problems (like the 30-60-90 triangle, the exponent substitutions you used here, etc...) would be amazing! Thanks!
Easy method here
Just take log on both sides and multiply it in side then you should have an equation with both sides having x of power one and substitute values of log40 ( log4 +log10) and log 2
Then u get 1.602x-1.602=0.601x+0.301
On bringing x oneside and constant on other side u get
1.001x=1.903
Which is approx answer x=1.903
If you need more precise then increase the decimal numbers
That's it not even a 2min task
That's what i was thinking it took me about 1 min to solve this mentally i was like he must have asked a numerical value i solved log 8 ans 3log2 = 3X.30 and i got 1.90
I arrived just after studying logarithms. And now this time it is again equality more exciting as always.
This is the first time I watch your videos and honestly they're actually helping me understand math even better, I don't know, I try to resolve the problem a little faster than you just to see if I can do it, sometimes I get it wrong but it's fine, I keep trying, thank you! you're helping someone get better at math!
I usually have the answer in my head before he finishes the countdown.
...then I watch the video to see HIS process and what the right answer is.
Of course it’s because we have 10 fingers, not because we count in base 10. I see we’re skipping the intermediate steps now
2:31 I think explaining that with the inverse function is more logical. Just like in _tan x = 2_ we take the inverse function of _tan_ on both sides, which is _tan^-1_ that gives us _tan^-1 (tan x) = tan^-1 (2)_ which simplifies to _x = tan^-1 (2)_
Here, the inverse function of _10^x_ is _log 10 (x)_ Taking that on both sides gives us _log 10 (10^x) = log 10 (80)_ which simplifies to _x = log 10 (80)_
I should start writing down these equivalences.
I did this problem a whole different way and still got 1.903 as a answer
Kinda boring, but you can solve this almost instantly using log(a^b)=b*log(a)
Yeah that’s what I was taught but I guess the way he does it can be understood better by most people
@@sauten2565 true and it also demonstrates far more, useful rules.
@@RealReen fax, how far have you gone in maths education, like A levels or further?
@@sauten2565 I'm not sure how it applies to your levels, but danish A-level, and I will be having a lot of maths a uni soon.
Hey Andy, unsure if you check the comments, but what software do you use to animate your videos? Thanks!
He said that a very few videos ago.
yeah, log can work. I just prefer natural log because it looks easier to format in my opinion.
But when the base is already 10…
How (and more importantly why) would you crowbar ln into this?
@@rathersane You can use logarithm regardless but yes, it's usually 10 as base. In this case, it works better. I'm just saying what I prefer in my opinion. Is that bad?
@@Grizzly01-vr4pn Crowbar? Huh???
@@Sg190th No… It just seems like more work in this particular case.
Hi man! What an amazing video! But, the verification can be done like this without a calculator:
40^(x - 1) = 40^(log80 - 1) = 40^(log8 + log10 - 1) = 40^(log8 + 1 - 1) = 40^log8
2^(2x + 1) = 2^(2log80 + 1) = 2^(2(log8 + log10) + 1) = 2^(2(log8 + 1) + 1) = 2^(2log8 + 2 + 1) = 2^(2log8 + 3) = 2^2log8 * 2^3 = 4^log8 * 8 = 4^log8 * 10^log8 = (4*10)^log8 = 40^(log8)
here, we can see that 40^(x - 1) gave us 40^log8 and 2^(2x + 1) also gave us 40^log8. so, verified.
What a smart and simple way to solve it by rules of indices 😻
Great video, it randomly appeared in my recommendations and i'm having a test on this exact stuff this week
This could be much simpiler. Use a logarithm to find what exponent makes 2 = 40. Expand that through x-1. Use algebra to find x.
I solved it in seconds just multiplying entire equation with log10 and got answer without calculator. Anyone with log knowledge can do it.
I knew that because I brute forced and put 1 as x and 2 as x and then just did 1.5 and kept raising it to 1.9.... They need to make it a bigger x so that I can't just input numbers quicker than it takes to actually use the correct formulas😅
You can also be silly and just find 2 to the power of what is 40 (comes out to 5.2319) and then finding x
“Log 10 means the same thing as log, and the reason for that is we have 10 fingers”
Wh… what?
Everything is base 10 unless otherwise stated. I'm not sure if our amt of fingers is the reason for that, but maybe it is
I checked this equation with gpt, the first answer for x=4. Then I asked to recheck and it was able to find the same answer but using log2
It'd be cool for you to make videos about our ten fingers and base 10 number system versus others like base 12.
I'm sure those too would be exciting.
I got ln(80)/ln(10) but it came out as the exact same thing, beauty of maths, different kinda methods but same answer if you do it right.
Quite silly, could have applied log base 2 at the outset and just solved thelinear equation.
if u try to compare the exponential powers it gives us the result as 2 which probably is a good approximation
pls correct me if i'm wrong thanks :)
I'm used to the ones that share a common base
would it be incorrect to just take the root of x-1 to both sides, which means the 40↑x-1 becomes 40, and the 2↑2x-1 becomes 2↑x and solve from there?
It took a minute for me to slove. Aaah.....
x to the a to the b. Whole thing squared.
if you apply significant digits - the answers are the same
as a ninth grader, anything after 2:31 is witchcraft
Very amazing 🎉🎉🎉🎉
Yay!!! I do it before of seeing the video and I got it right🎉🎉🎉
i used the natural log from the first step
As an Asian parent I would say this should be easy for my 2 year old one
Fascinating ….!
Doing this myself before watching the video was weird, it didn't feel right to get such a nice result.
Because we have 10 fingers! 😂
''its because we have 10 fingers''
How exciting🗣🗣🔥🔥🔥
Indeed that is how is supposed to be done, 👍
Without using log i some how by unitary method only i got two values of x which are 1 and 4 and this not possible and my method was also not wrong.
You could have done a 2nd level unknown equation to get it without a calculator
Lol i sloved it in my head 😅😅
Take log in the first step;
(x-1)(2log 2 + log 10) = (2x+1) log 2
=> 2x log 2 - 2log 2 + x -1 = 2x log 2 + log 2
=> x - 1 = 3(log 2)
=> x - 1 = 3(0.30103) => x = 1.90309
That's more what I was thinking lol, how they show us in school. xd
I solved it in another way and i found the same result as yours, i am more happy to send it to you if you want to see how
im so happy that i could figure this one out in the thumbnail itself heh yay
just introduce logs to both sides in the beginning
40^x-1=2^2x+1
log 40^x-1= log 2^2x+1
power law of logs, the power of the log can become the coefficient of the log
(x-1) log 40 = (2x+1) log 2
x log40 - log40 = 2x log2 + log2
move the x terms to the left, and the log40 to the right
x log40 - 2x log2 = log 2 + log 40
factorise x from the right
x(log 40 -2log2) = log2 + log40
x= (log2 + log40) / (log40 -2log2) = 1.903089987
why didn't you just ln the both sides from the first ? it would've been much easier.
x=1.9 =>
(x-1)log40 = (2x+1)log2
wolfram here we come
Andy!!
Решил по формуле Пика за 0,000000002 секунды
I thought it was because we have 10 toes
much exiting
There was a really easier way, you could write as (2^ 5.32192809490)^x-1=2^2x+1 ,since both bases are the same we can write down as 5.3219280949(x-1)=2x+1 which is then equal to 5.3219280949x-2x= 5.3219280949+1 then x= 6.3219280949/3.3219280949
Why i always want to take log😂
How exciting.
I'm in 10th class from imdia and solved it very easily
i’m in 6th from the usa and solved it really easily
so simple you just know about log 5on2 =2.3 and done and solve it
but you can do it simple you should change this log to log5 /log2 => 0.7/0.3 =2.3 (rounded )
and you dont wana calculator or somthing this question was standard for student
Shouldn't a and b from the LHS and RHS be same?
Nice
could you also just do x-1(log 40)=2x+1(log 2) and solve that way?
actually It's much easier that way
Log 40 could be written ans log 4+log10.The only thing is we should know the value of log4 and log2.
amazing thank you
in the USSR, children solved such examples in nurseries
IMO childrens thinking capabilities way more powerful. Age won't matter, IG time does.
Approximately 2, right?
Is nobody going to say how easily it could have been solved by using log
Couldnt u do on comparing powers?
@andymath what kind of training do you do for your deep voice?
I got it right somehow 💀
i got x = 1 + 3log2
Log 2 equals 0.30 and 3 times log2 is 0.90 adding one to it will result in 1.90....the answer is same
Thank you bro
iIs the reason for that really that we have ten fingers??
Thank you, it is actually really funny !!!!!!!!!!!!
From 1st sec, i striaght knows that just using ln() then find the x. Just that simple. No need to be simplify🤔 why need sinplify as we can just using ln() loll
that too easy with 16 years old vietnamese people =))
Can you give me an example of how this is useful outside of math?
problem solving is a very important transferable skill. Being good at solving any math problem means your likely a good problem solver in any setting. Yes, you won't need to solve this specific equation or anything like this in a practical setting, but it trains your brain to think outside the box
@@satyadas4751Your comment looks important, let's put a box around it.
It’s an exponential function. You can google the purpose and application of those. This is a problem solving channel, not an application and vague question answering channel.
it's exciting
that's like asking how is breathing useful outside of respiration.
I got 1.8 😢
No base drop = thumbs down
(Ln80)/(Ln10) gang
i knew the answer would be in terms of log, so kinda disappointing
It's log more exciting
why don't you apply LOGARITHM on both sides
Did he do this one right?
by trial and error i found 1 then 2 then 1.9. and 1.9 is way way too close. I found in 3rd iteration via calculator. I will take that as ok becuz in real life that would negligible difference, mostly covered by buffer and in real life u won't be fucking sitting there finding value of log 69 or something. Maths doesn't has real life applications beyond class 10 or 12 .
Edit: the answer is 1.90309 lol
That's 0.16 % difference.
Can I send you a problem?