We solve this exponential to a non-integer power by introducing a new operator.... then reversing it's output by solving a new number raised to a new non-integer power. While correct, this is still thoroughly cursed recursion. "Look it up in tables" is not a satisfying answer.
I'm not sure changing 2.1^3.37 to e^2.50032 simplifies much of anything. Sure now you can reference a book that lists powers of e, but how were they obtained? This is a circular solution.
I agree. If the point is to learn the underlying math and not rely on a calculator then what are we using lookup tables for? That's no different from using a calculator. Here is how you can actually do the work by hand. First, to calculate ln(2.1) solve for u such that 2.1 = (1 + u)/(1 - u). The formula is u = (2.1 - 1)/(2.1 + 1) = 11/31. Then use the series expansion ln(2.1) = ln((1 + u)/(1 - u)) = 2(u + u^3/3 + u^5/5 + ... + u^(2n-1)/(2n-1) + ...) Then, to calculate e^2.50032 you can use the famous series e^x = 1 + x + x^2/2 + x^3/6 + ... + x^n/n! + ... By the way, the whole point of using base e is that we get these lovely series expansions. I don't know why base e was used in the video since the log table ended being in base 10.
It is circular in that sense, but I think he was just trying to get around the overflow error when evaluating 2^337, not completely avoid using a calculator. Although the title makes it seem like he is teaching how to calculate this without a calculator which he isn’t
@@martinepstein9826 Yes, it’s circular when it comes to the exponent action part. Because then you are left with e^2.50032 and then you are in the same situation that you started in. The only way to calculate this is to use a calculator. Which is fine if that’s what he said in the beginning because he used this trick to circumvent the 2^377 overflow problem.
I was on tenterhooks during the first half of the video, wondering whether you were going to use logarithms, or whether you were using a flashy new method that I hadn't heard about. You're right though. Logarithms were an extremely useful tool which were sensational when Napier and others invented and improved them during the 17th century. They were a very important tool for those mathematicians and scientists who were in the process of developing 'the scientific method'.
Good point about the sensation logarithms must have caused at the time. Similar to the revolution that electronic computers have brought about in our age.
Didnt answer my question though. Just ended with e to the power of another decimal which is where we started basically. How would we find the answer without a calculator. We started with a number to a decimal and ended with the same thing but more complicated.
But then, how do early people calculated the values in the log tables in the first place? Also, why you still used a calculator when the title of the video says NOT using a calculator?
How nostalgic to bring back my college math, physics, and chemistry 27 years ago. I'm now a physician and had totally forgotten these topics. Thanks for bringing back my memories. If Euler and Newton are alive right now, they will give you many THUMBS UP.
Good vibe from Morelia. Now I have the two explanations, discovermaths and shurprofe. Continuing with the channel marathon in English and it is ready. Let's go for it.
Sir I thought you will tell any traditional values to calculate but at the end you had used the calculator. If we have to use the calculator then why not simply get the answer by putting 2.1^(3.37). Sorry to say this sir but if you have to calculate the value of exponential and natural log by using calculator then please change the title of the video. I'm really sorry if I have disrespected your sentiments.
Cool video, but you having to raise e to something with a decimal ruins the whole point imo. If you're going to use a calculator at the end. Might aswell have just done the first calculation with a calculator.
I don't understand the "multiply it by itself so many times" idea. 5 multiplied by itself is 25 - we multiplied it by itself once. Why claim we multiplied it by itself twice?
how can you solve it by hand when you still ending up with e^2,50003 ... that is decimal number as well so still same problem but different numbers :DD
If you can't use logarithms, use the hint from the beginning of the video. 4^.4 = 4^(2/5) = 16^(1/5) 4th root of 16 is 2, so the 5th root is going to be less than that. Guess 1.7, take it to the 5th power and see what you get - too low. Repeat with 1.8, too high. Guess again with 1.75 - high but better. 1.74 is close and you only need a little more. Keep it for as long or as much precision as you need. Or, get a calculator. By the time you need to solve things like 4^.4 you're going to need one.
In the end we should use a calculator to get the answer !!! And one more thing !! did you relate to Bill Nighy by any chance ?? you look like him and your voice sounds like him !! Thanks for explanation
yeah and how are the values in the log table calculated? calculator didn't exist back then. Tell how to calculate the exponent truely by just our hand and knowledge
Of course we can find the answer without a log table. Someone had to create the log table in the first place, didn't they? Sadly, the video doesn't explain how to do this but you can use the series expansions ln((1 + u)/(1 - u)) = 2(u + u^3/3 + u^5/5 + ... + u^(2n-1)/(2n-1) + ...) e^x = 1 + x + x^2 + x^3/6 + ... + x^n/n! + ... Personally, I don't find solving with a log table any more interesting that solving with a calculator.
@@martinepstein9826 e^2.50032= e^2 (e^.5) (e^.00032) e^2 and e^.5 are solvable by hand calculations as long as you know e=2.71828182846 as for e^.00032 is approximated by (2+.00032)/(2-.00032) =1.0003200512 accurate to 12 digits.
@@MathTidbits I was thinking about how to get a similar approximation for ln(1+x). We can just invert your formula: 1 + x = e^y ~= (2+y)/(2-y) y ~= 2x/(2+x) Again, for small x the difference is about x^3/12
A calculator isn't more lazy than just reading a log table or a slide ruler. 'real maths', if that's what we are going to call it, would be to approximately find the solution using something like CORDIC or some numerical method.
maybe im dumb but i came very close in my head of that answer in 8 or 9 sec just by doing 2 exponant 3 en then i tell my self what is 1 / 2 between 2 exponant 3 and 2 exponant 4
"and thats how logarithsms... were.... obtained... in the past..." ah yes, the logarithm tables given to us mortals by the gods! ... ahhh sigh, basically you just told me you dont really know how any of this works, because you dont even know where the numbers in the log tables come from :/ good riddance mate.
![The Most Useful Curve in Mathematics [Logarithms]](/img/n.gif)
Wait hold up and how do you calculate e^2.500… without a calculator 🤣 Bru
Base conversion
@@Kashisulu I see, thank you
@@Kashisulu can you explain
@@shabeerp1153 th-cam.com/video/Z2AY-bWsezk/w-d-xo.html
Refer
@@Kashisulu 10^(2.500… log10(e))
We solve this exponential to a non-integer power by introducing a new operator.... then reversing it's output by solving a new number raised to a new non-integer power.
While correct, this is still thoroughly cursed recursion. "Look it up in tables" is not a satisfying answer.
Your videos are always interesting and worth watching. It makes me enthusiastic to do more maths.
Hi
I'm not sure changing 2.1^3.37 to e^2.50032 simplifies much of anything. Sure now you can reference a book that lists powers of e, but how were they obtained? This is a circular solution.
yeah exactly then how to calculate e^2.50032 😂😂
I agree. If the point is to learn the underlying math and not rely on a calculator then what are we using lookup tables for? That's no different from using a calculator. Here is how you can actually do the work by hand.
First, to calculate ln(2.1) solve for u such that 2.1 = (1 + u)/(1 - u). The formula is u = (2.1 - 1)/(2.1 + 1) = 11/31. Then use the series expansion
ln(2.1) = ln((1 + u)/(1 - u)) = 2(u + u^3/3 + u^5/5 + ... + u^(2n-1)/(2n-1) + ...)
Then, to calculate e^2.50032 you can use the famous series e^x = 1 + x + x^2/2 + x^3/6 + ... + x^n/n! + ...
By the way, the whole point of using base e is that we get these lovely series expansions. I don't know why base e was used in the video since the log table ended being in base 10.
It is circular in that sense, but I think he was just trying to get around the overflow error when evaluating 2^337, not completely avoid using a calculator. Although the title makes it seem like he is teaching how to calculate this without a calculator which he isn’t
@@Tony-cm8lg It's not really circular. You just need to look up in the log table which number has a log of 2.50032
@@martinepstein9826 Yes, it’s circular when it comes to the exponent action part. Because then you are left with e^2.50032 and then you are in the same situation that you started in. The only way to calculate this is to use a calculator. Which is fine if that’s what he said in the beginning because he used this trick to circumvent the 2^377 overflow problem.
I was on tenterhooks during the first half of the video, wondering whether you were going to use logarithms, or whether you were using a flashy new method that I hadn't heard about. You're right though. Logarithms were an extremely useful tool which were sensational when Napier and others invented and improved them during the 17th century. They were a very important tool for those mathematicians and scientists who were in the process of developing 'the scientific method'.
Good point about the sensation logarithms must have caused at the time. Similar to the revolution that electronic computers have brought about in our age.
Didnt answer my question though. Just ended with e to the power of another decimal which is where we started basically. How would we find the answer without a calculator. We started with a number to a decimal and ended with the same thing but more complicated.
Use maclaurin series of e^x
@@h1m4n8hu Whats that
e^x = lim n→∞ (1+x/n)ⁿ
That can be calculated by hand.
thank you! this helped me with an antilog problem
Good explanation. How were the log tables set up?
Can we solve this without log table and calculator?
But then, how do early people calculated the values in the log tables in the first place? Also, why you still used a calculator when the title of the video says NOT using a calculator?
Great finally I found what I want
me too
Me also
I too
How nostalgic to bring back my college math, physics, and chemistry 27 years ago. I'm now a physician and had totally forgotten these topics. Thanks for bringing back my memories. If Euler and Newton are alive right now, they will give you many THUMBS UP.
Physician or Physicist
@@Sougata_XD kid named finger
Wait, you can solve e^2.50... in mind but you cant solve 2.1^3.37🤣🤣🤣
Good vibe from Morelia. Now I have the two explanations, discovermaths and shurprofe. Continuing with the channel marathon in English and it is ready. Let's go for it.
very useful for calculating pH (log) in chem #jee2023
I had forgotten this method and searched a lot for it.
Finally got it.
Thanks man
What is the result at last.... You again got Stuck in solving exponential power
You finally made it make sense. Thank you!
This video is absolute treasure.
Thank you sir.
A slide rule is a calculator!
Indeed! Perhaps we should have specified "electronic calculator".
Do the 100th root of 2.1 first and then raise it to the 337 power!
what? it dont make sense
2.1^3.37 it's hard to solve
but e^2.500328852 isn't more easy to solve '-'
e^2.500328852 is not hard to solve;only impractical.
e^2.500328852 ~ (e^2) * (e^.5) * ( 1 + .000328852)
doable by hand calculations,but too tedious.
Lol in the end all you did was raise 2.71 to 2.5th power... It's the same thing again... Can't do it without calculators
Awesome tutorials,learned a lot, clear and precise
Can you help me to how to find the value of exponential power function?
For example : e^1.8=??
Use antilog table
Use the series e^x = 1 + x + x^2/2 + x^3/6 + ... + x^n/n! + ...
Use scientific calculator 😊
I needed to be able to do such calculations 100% without calculator. Is that possible cuz mcat seems to think so
Sir I thought you will tell any traditional values to calculate but at the end you had used the calculator.
If we have to use the calculator then why not simply get the answer by putting 2.1^(3.37).
Sorry to say this sir but if you have to calculate the value of exponential and natural log by using calculator then please change the title of the video.
I'm really sorry if I have disrespected your sentiments.
Thanks man it worked😊😊😊😊 but how did you calculate e^2500... without a calculator?
Thank you soo muchh... Really helped mee
U didn’t teach how to calculate the number though, it was depend on the calculator like ln2.1 and e^2.5
Cool video, but you having to raise e to something with a decimal ruins the whole point imo. If you're going to use a calculator at the end. Might aswell have just done the first calculation with a calculator.
💗thanks man learned something new and helpful
*_Then what will happen with negative base to decimal exponent?_*
But you used a calculator...
Sir can u make me brilliant at maths
We'll do our best. Keep watching the videos!
so, what if I do not have a book???
I don't understand the "multiply it by itself so many times" idea.
5 multiplied by itself is 25 - we multiplied it by itself once. Why claim we multiplied it by itself twice?
After wasting 3hours in other videos I finally found this.. Thanks!!!
but then how do u compute the last e^big decimal?
By antilog table dude
Sir Can you please make a video on finding powers without decimals
how can you solve it by hand when you still ending up with e^2,50003 ... that is decimal number as well so still same problem but different numbers :DD
that's what i was thinking too lol
Only 2.5
well, you could reasonably use 5/2 as an approximation and solve it that way
@عبدالله tryeasy rewrite it using exponent properties first. 2^2 * 2^1/2
Amazing editing + amazing content
there is someway to do that without ANY calculator? Because it was used to calculate ln 2.1? please, I need to find a way to calculate this
Now,how logarithms were discovered
thank you sir it really helped me a lot
Thank you so much for this wonderful explanation😊😊
Thanks!!!!
Now tell me using with calculator
You did not explain "e"
I always wanted to see how a number half of its times would look without calculating. I wish logarithm didn't exist.
How do u find log of 2.1
But we have to use calculator to calculate e^x
nope, he explains that you need to use log tables in 'reverse' @5:56
wow this is truly above my league, as I don´t know how to work with logs and ln´s. But, what about evaluating 4^0.4 without a calculator?
Take log of 4. Multiply by 0.4 . Take antilog of that result.
If you can't use logarithms, use the hint from the beginning of the video.
4^.4 = 4^(2/5) = 16^(1/5)
4th root of 16 is 2, so the 5th root is going to be less than that. Guess 1.7, take it to the 5th power and see what you get - too low. Repeat with 1.8, too high. Guess again with 1.75 - high but better. 1.74 is close and you only need a little more. Keep it for as long or as much precision as you need. Or, get a calculator. By the time you need to solve things like 4^.4 you're going to need one.
In the end we should use a calculator to get the answer !!!
And one more thing !! did you relate to
Bill Nighy by any chance ?? you look like him and your voice sounds like him !!
Thanks for explanation
U said we will not use cal. Yet to didnt explain how you got log values from...(id we dont hv log table or any other artificial means)...😒😒😒😒
Ah log tables! I used to own a set of these...
I kept mine from school - half a century ago!
@@discovermaths Sadly my copy disappeared somewhere during one of my many house moves over the years.
At 2:24 ... never related so much with a software. I too can't work in this range:(((
Very helpful..thank you sir
I understand why there's so many dislikes 🤣
If there is anything written in addition or subtraction with given ques then what to do please explain
Lastly he used calculator😂😂😂😂😂😂😂
Used a calculator. False advertising.
yeah and how are the values in the log table calculated? calculator didn't exist back then. Tell how to calculate the exponent truely by just our hand and knowledge
Could you show how to do trigeometry without calculator
I still can't process how fractional, non-whole exponents are possible. Seems counter intuitive at first.
So we cant find answer unless we have logarithmic table. Is it??
Of course we can find the answer without a log table. Someone had to create the log table in the first place, didn't they? Sadly, the video doesn't explain how to do this but you can use the series expansions
ln((1 + u)/(1 - u)) = 2(u + u^3/3 + u^5/5 + ... + u^(2n-1)/(2n-1) + ...)
e^x = 1 + x + x^2 + x^3/6 + ... + x^n/n! + ...
Personally, I don't find solving with a log table any more interesting that solving with a calculator.
@@martinepstein9826 e^2.50032= e^2 (e^.5) (e^.00032)
e^2 and e^.5 are solvable by hand calculations as long as you know e=2.71828182846
as for e^.00032 is approximated by (2+.00032)/(2-.00032) =1.0003200512
accurate to 12 digits.
@@MathTidbits I've never heard of (2+x)/(2 - x) being used as an approximation for e^x. The difference is only about x^3/12. Pretty clever.
@@martinepstein9826 I stumbled upon this by modifiying Newton method of approx.
Limit as N approaches infinity [(N+1)/(N-1)]^(N/2) approaches " e "
@@MathTidbits I was thinking about how to get a similar approximation for ln(1+x). We can just invert your formula:
1 + x = e^y ~= (2+y)/(2-y)
y ~= 2x/(2+x)
Again, for small x the difference is about x^3/12
Sadly this can't be used by a 8th grader. I thought I'd be easier
it doesn't work for me sorry
A calculator isn't more lazy than just reading a log table or a slide ruler. 'real maths', if that's what we are going to call it, would be to approximately find the solution using something like CORDIC or some numerical method.
How to calculate the approx value of base 10 with power in decimal???
Please reply soon🙏
Sir i knew it but this trick is not going to work in compatative exams.😅
India study for their stomach
@@sarthaksrthkyoutu100 yeah buddy
For stomach
maybe im dumb but i came very close in my head of that answer in 8 or 9 sec just by doing 2 exponant 3 en then i tell my self what is 1 / 2 between 2 exponant 3 and 2 exponant 4
How did you jump to 12.185 from 2.5, i did not get that.
There is an antilog table just like a log table, for performing the inverse function of log. If you look up on that, you get antilog of 2.5 is 12.185
You set that as the exponent and use base e
how did yuo get .7419 for ln 2.1 without a calculator
How come when you do 0.517 to the power of 1.3 the answer is 0.424? Surely if your multiying something larger than 1.0 it will be bigger not smaller??
It’s because you can write it as 0.517^1 * 0.517^0.3. 0.517^0.3 is about 0.81 which * 0.517 means you get 81% 0.517 or about 0.435
0.517 is not greater than 1
Thanks a lot Sir
Awesome thanks alot sir
Thanks sir
Logarithem is not allowed school then to now how what shall we do in India
Didn't he use a calculator after all?
Thnx sir 🙏
I'm 12 years and i didn't understood anything 😭
😂😂 what a great calculator
2.1 ^3.37 using a simple calculator?
Can anyone do it?
Without log table
Ah i thought u did without using log
thanks!!
No no it's beyond my head
"and thats how logarithsms... were.... obtained... in the past..." ah yes, the logarithm tables given to us mortals by the gods! ... ahhh sigh, basically you just told me you dont really know how any of this works, because you dont even know where the numbers in the log tables come from :/ good riddance mate.
best video
You started too late on TH-cam,,, Master 😭
What about this , ⁴√³√2²
Use differentiation
l am 8th grade and l memorized it
Wow! Thanks, I got it now.
💙
Waste
Again e power of decimal
Why he all the time smile and make sound by his lip and tounge , he get on my nerve
It is very much irritating