Solving A Quick and Easy Logarithmic Equation
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- Опубликовано: 14 янв 2022
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It’s enough to know this identity a^log b = b^log a. And you can solve it in seconds.
Exactly. İ did it in this way , so much better
nver knew that there was such iidentity. Still managed to solve it though. Thanks for the info
But isn't that incorrect? Please check it. A power b is not the same as b power a
@@rajasekhark8333 yes but with the "log" it is the same , you can check
@@rajasekhark8333 a^logb=b^loga is correct,but a^b is not = to b^a
Alternative method:-
a^log b=b^log a
Here, x^log 25=25^log x=k
given, x^log 25+25^log x=10
k+k=10,
k=5
Now compare 25^log x=5
(5)² ^ log x=5
5^log x²=5
log x²=1
x²=10
x=√10.
Thank you ☺☺
In 1st method when you have 10^(y*log 25) you could simply make (10^log 25)^y, which turns to 25^y. You would then have 25^y + 25^y = 10, which leads to y=1/2. Just a shortcut. Anyway great video
That's right! Thank you!
Hey, when you have 10^logz + z = 10, you do not need a substitution for t, because 10^logz is just z. So, 2z=10, z=5 straight. Another substitution leads to final result, but it is just like z=t after all. But cool problem, I wasn't trying to solve it. : )
Using identity x^(log_b n) = n^(log_b x) makes it really easy!
Basically identity says both terms are the same. They you have two methods, use x in the base or in the exponent.
1st method:
x^log 25 + 25^log x = 25^log x + 25^log x = 2 * 25^log x = 10 => 5^2.log x = 5 => x=sqrt(10)
2nd method:
x^log 25 + 25^log x = x^log 25 + x^log 25 = 2 * x^25 x = 10 => x^log25 = 5 => ... =>x=sqrt(10)
Right!
The essence of solving this problem is realizing that the two terms on left side are equal.
a^log(b) and b^log(a) are equal since their logs are log(b).log(a) = log(a).log(b). Therefore each term is 10/2 = 5. Taking log, we get -
log(5) = log(x)log(25 = 5^2) = 2.log(x).log(5) => log(x) = 1/2 => x = 10^.5 = sqrt(10)
?
Funny, I saw the same thing. Before even starting the problem, I played around a little with the idea of taking the logarithm of both quantities on the left side of the problem equation, I was kinda brainstorming and kinda daydreaming. I realized that log(x^(log(25)) and log(25^(log(x))) both yield log(x)*log(25), so they're equivalent. log(x)*log(25) = log(5) is a MUCH easier problem to solve.
The first method reminded me of the film inception, a replacement within a replacement within another one, very interesting. Nice video as always.
Thank you! 💖
Interesting question and nice solutions. Thanks for sharing!
Np. Thank you! 😊
It's a standard log property that x^log y = y^log x.
Proof: x^log_b y = (b^log_b x)^log_b y = b^(log_b x * log_b y) = (b^log_b y)^log_b x = y^log_b x
This one was fun, just saw it. Good work.
Thank you!
"I'll be presenting two methods - let's start with the first one." Impeccable logic!
😁
@@SyberMath As a retired maths teacher, I always did my best to inject humour into my lessons and show that maths can be fun. I enjoy your presentations - they bring back memories.
Another method besides what I saw from here is to use the properties of exponent/logarithm using the base and exponent to switch. The equation turns into 25^log(x)+25^log(x)=10. This means that 2(25^log(x))=10. Dividing both sides by 2 gives you 25^log(x)=5. Turn the left side of the equation as (5^2)^log(x)=5. Since the base are the same, then the exponents are equal to each other. Using properties of exponent, 2log(x)=1. Dividing both sides by 2 is going to be log(x)=1/2. Therefore, x=10^(1/2)=√(10).
It took you (and me, too) just 30 secs to solve, but him... 9:26...
Amazing video as always!
Glad you enjoyed it!
Let log(x) = y, then x = 10^y.
So (10^y)^log(25) + 25^y = 10
10^[log(25)*y] + 25^y = 10
10^log(25^y) + 25^y = 10
25^y + 25^y = 10
25^y = 5, which means y = 1/2.
But x = 10^y, so x = 10^(1/2) = sqrt(10)
Nice.
Thanks for this logarithmic equation :) i just like this kind of equations )) thanks for your videos and solutions ))
Oh np! Thank you! 💖
0:55 i totally agree with you here
a ^log (b) = b ^log (a), which means 25 log x + 25 log x = 10 25 log x = 5 , log x = 1/2 since 25^1/2 = 5, and then we ge t the x = 10 ^1/2
Two Z or not two z! Hillarious! I like that you comment it as if you were just solving it as you speak instead of choosing the more professor-like approach…!
Thank you!
U'r amazing with ur multiple methods
Thank you! 💖
Awesome!
Thanks!
Hey there, i'm a big fan of your videos, and i've been watching for a while now, but i was wondering if you could maybe do one in how to solve 2^n > 2n+1
Take the derivative.
thank you for your amazing video!
Np. Thank you! 💖
a^log(b) → log(b).log(a) →b^log(a)
Gives the general identity:
a^log(b) = b^log(a)
I really liked the methods and there is also a good and neat property of logarithm says that a^log base b of c. =c^log base b of a ,,, which can also solve this problem
Absolutely! That would be a good way to approach it!
Very very very clever!!!
Thank you!
what a problem!!!! very interesting
After watching a couple of your videos
I got to know that you're a big big fan of Substituting terms with other variables
yess!
Thank you for your great videos...
And I want to ask you:
I found from wolfram alpha that
sum of k^(-k) from k=1 to k=infinity is: 1.29129 but why?
So Unique and doesnt even have a close form except for 129129/10000
Show de bola!!!
10^(log(z)) = z. What was that long detour?
All videos > 8 min can have midroll ads and that boosts his revenue.
excellent 😃
Thank you! 💖
Over 40k views this one, the thumbnail s working great!
Using the property
a^log(b) = b^log(a) , you could write the equation as
25^logx + 25^logx = 10
=> 25^logx = 5 = 25^(1/2)
=> logx = 1/2
=> x = root(10)
:D
Some people write log to mean log to base e (which is also denoted as ln)
Nice,well done!(10^logx)^log25+25^logx=10
(10^log25)^logx+25^logx=10,5^(2logx)=5,2logx=1 so logx=0 5 and x=sqrt10.
log(x^log(y))=log(x)*log(y)=log(y^(log(x)) => x^log(y)=y^log(x), for log to any base. Ha ha, very nice misdirection, why did I find it so hard?
Great, my congratulations! Can you make an video explaining how to solve Integral (tang(x))^(1/2)dx please?
Hi, thanks! You're not the first person to ask that! 😁
Here's a video that @blackpenredpen made:
ruclips.net/video/6v0QqJ87gdM/видео.html
I thought both these methods were more complicated than they needed to be. Write x^(log25) = 10^[log(x^(log25))] = 10^(log25 logx) = 25^(logx) and then finish it off.
I agree! 😁
10^(log(z)) = z. What was that long detour for?
nice solution
Thank you, Nico!
Fantastic
Thank you so much 😀
congrats for your 70k btw
Thank you so much 😀
I think a third method would be the following:
let 10 = 5 + 25^(1/2)
x^(log(25)) + 25^(log(x)) = 5 + 25^(1/2)
then I can solve for each term:
x^(log(25)) = 5
x = 5^(1/log(25))
25^log(x) = 25^(1/2)
log(x)=1/2
x = 10^(1/2)
btw I like your videos they're amazing
Good thinking! Thanks
How can you assume two different numbers equal to t. ?
They are not different
Hi Syb, instead of using log base 10, why not trying log base 25 ? If I am right, this should give a nicer result.
Besides that, thank you for this very cool channel.
That's right! Thank you!
Thx
No problem
As everyone said, I solved it in 6 seconds.
It's easy to know that 25^logx=x^log25
So 25^logx=5
logx=1/2
x=(∛1000)^0.5
Both methods are nice.
Thanks a lot
Moreover, you can notice that on the left function is increasing everywhere. So you can find a solution yourself. This is my the third method.
Thank you very much!
you must be happy with all of these substitutions (i hope so)
Yess! 😍
Sqrt(10) - easy guess and monotonicity finishes the proof
√10
If a>0 b>0 then a^logb=b^loga x^log25+25^logx=10 2×25^logx=10 25^logx=5 logx ×log 5 25=1 logx=1/2 x=sqrt10
set x^log25=t log25*logx=logt so logx=logt/log25 25^(logt/log25)=t so 2t=10 t=5 so x^log25=5 log25*logx=log5 log x= 1/2 x=10^(1/2)
Isn’t 10^logZ is equal to simple Z?
yes
I was taught in school that log(10)x=lgx and logx is a syntax error
Some countries use different notations unfortunately.
x^log(25) + 25^log(x) = 10
Note 1: log[x^log(25)] = log(x)*log(25) = log(x)*2*log(5)
Note 2: log[25^log(x)] = log(25)*log(x) = 2*log(5)*log(x)
Both terms equal!
2*25^log(x) = 10
25^log(x) = 5
log(x) = 1/2
x = sqrt(10)
Also ...
x^log(25) + 25^log(x) = 10
With log(x) in equation, we know x > 0.
Thus, we know 10^log(x) = x . Substituting, we have:
(10^log(x))^log(25) + 25^log(x) = 10
10^(log(x)*log(25)) + 25^log(x) = 10
10^(log(25)*log(x)) + 25^log(x) = 10
(10^log(25))^log(x) + 25^log(x) = 10
25^log(x) + 25^log(x) = 10
25^log(x) = 5
5^(2*log(x)) = 5
2*log(x)) = 1
log(x) = 1/2
x = sqrt(10)
One of Newton's best works.
Another perfect video.
Wow, thank you! 😍
Noice!!! Sybermath
May this channel reach 1M subscribers soon!!!
Thank you! 💖😊
Just watched this video for the second time
Amazing!!!
Thank you!!! 🙂
🎉❤️❤️❤️
💖😊
X=sqrt(e)
log is base 10 😉😁
x=√10
why?
@@SyberMath x^log25=m
log (x^log25)=logm
log25*logx=logm. ______1
25^logx=n
log (25^logx)=log n
log x*log 25=log n
This shows m=n
2*25^logx=10
25^logx=5
(5^2)^logx=5
5^2logx=5
2logx=1
logx=1/2
x=10^12=√10
*"Logo-exponential"*
Sounds like a dog breed
.nice🙏🙏
x^(log 25) + x^(log 25) = 10
2 (x^(log 25)) = 10
x^(log 25) = 5
log x^(log 25) = log 5
(log 25)(log x) = log 5
log x = log 5 / log 25
log x = log 5 / log 5²
log x = log 5 / 2 log 5
log x = ½
x = 10^(½)
x = √10
If you told log with base 25 it's easier 😂 : x = 5
That's right, Damien!
your videos are very informative. but you can make them much better using only m and n as variables. in order to avoid these annoying singings and puns about be, are, to you, why, tea and coffee. they're not useful and they neither gain sense by endless repetition.
aside of that i like the videos very much!
Noted
好，奖 1 万元