I think the simplest approach is to use the change of base formula, but applied only to the right-hand side of the equation. Just convert it to log[5]25/log[5]x. Then cross-multiplying, etc., leads straight to the answer.
As soon as I saw the problem, I immediately went with the 3rd method, since it's practically the most obvious one. The other two methods seem unnecessarily convoluted to me.
So did I but I missed x = 5^(-sqrt(2)). Logarithms are very important for Analytical Chemistry and I, as a chemist, must teach students because our mathematicians usually miss this material.
Given: log_5 (x) = log_x (25) To find: x Using change-of-base: log(x)/log(5) = log(25)/log(x) As log(x) ≠ 0 by definition: (log(x))² = (log(25))·(log(5)) Using log(a²) = 2log(a): (log(x))² = 2·(log(5))·(log(5)) (log(x))² = 2·(log(5))² log(x) = ± √2·log(5) log(x) = log(5↑(±√2)) Taking both sides to the power of 10: x = 5↑(±√2)
If we logarithmize whith a base of 5, a shorter solution is obtained. LOG X(5) = LOG 25(X) , LOG X(5) = LOG 25(5) / LOG X(5) LOG X(5) . LOG X(5) = 2 .....
Convert the equation to natural logs:
lnx / ln5 = 2ln5 / lnx
(lnx)² = 2(ln5)²
x = 5^±√2
When using change of base, you can save some steps by changing everything to log base 5 instead of a generic log so you can keep the LHS as one log
I think the simplest approach is to use the change of base formula, but applied only to the right-hand side of the equation. Just convert it to log[5]25/log[5]x. Then cross-multiplying, etc., leads straight to the answer.
Agree
How?
Yeah...
I Changed the base
Easily i got the answer
I am not good at math, so I watch these videos to learn. Hopefully I get better.
I used change of base with 5 as the base. Led to the same result, 5^(+/-√2)
X=5^(sqrt(2)) or 5^(-sqrt(2))
It's a very simple equation for solving in head.That:
Logx/log5=log25/logx
or,(logx)^2=2(log5)^2
or,logx=√2log5,-√2log5
or,x=5^√2,5^-√2
or,x=
Thank you!!!
Best way to do it
Log 5 (x) = log x (25)
Log 5 (x) = 2 log x (5)
Log 5 (x) = 2 / log 5 (x)
[(og 5 (x)]² = 2
Log 5 (x) = 2
X = 5^√2
x=5^(±√2)
I did in mind in 10 seconds!!!
Wow! 😁🤩
@@SyberMath :D
that was great! 😍 i think i did a 4th method but pretty close to method 1. Awesome video
Awesome! Thank you!
Nice!
i naturally went straight to the 2nd method. i'm a "change of base" kinda guy.
As soon as I saw the problem, I immediately went with the 3rd method, since it's practically the most obvious one.
The other two methods seem unnecessarily convoluted to me.
I solved it in my head in a few seconds
me too, but i always forget the negative sqrt
😂😂😂😂
So did I but I missed x = 5^(-sqrt(2)). Logarithms are very important for Analytical Chemistry and I, as a chemist, must teach students because our mathematicians usually miss this material.
Me too
@Phoenix ... But how do u prove that to people ... I mean we are not inside ur head🙄...so just how true is ur claim?🤔 ... 😂😂😂
x = 5^(sqrt)
Given:
log_5 (x) = log_x (25)
To find:
x
Using change-of-base:
log(x)/log(5) = log(25)/log(x)
As log(x) ≠ 0 by definition:
(log(x))² = (log(25))·(log(5))
Using log(a²) = 2log(a):
(log(x))² = 2·(log(5))·(log(5))
(log(x))² = 2·(log(5))²
log(x) = ± √2·log(5)
log(x) = log(5↑(±√2))
Taking both sides to the power of 10:
x = 5↑(±√2)
hocam türkiyeden böyle bir içeriği ingilizce dilinde aktarmanıza bayıldım.. tebrik ederim
Tesekkur ederim! 🥰
@@SyberMath rica ederim hocam. Türkçe kanalınız varsa onu da takip etmek isterim
If we logarithmize whith a base of 5, a shorter solution is obtained.
LOG X(5) = LOG 25(X) , LOG X(5) = LOG 25(5) / LOG X(5)
LOG X(5) . LOG X(5) = 2 .....
Hello teacher 🙏
Hello! 🤩
There is a small error at the 2nd method
I made up this equation but I don't know how to solve it
Equation:
logx(5x)=2x/5
Answer: 5
Not any substitution needed,nor any reciprocal rule,flipping etc. The second method is simple and perfect as well
Good to hear!
I couldn't understand why 2log5's 2 become root 2 in next line's square...why not instead of 2???
it's 5^√2
Confusing!
Why? or what part?
X=5^(sqrt2),x=5^(-sqrt2)
y?
Next video: can you solve x=1?
Nope.
Solve x-1=0 or x-1=x 😁😜
@@SyberMath There are so many nice tasks and you are quite creative in them. Why do you select
trivial ones?
@@SyberMath For you from Sivashinsky text book
x^4+(x+1)(5x^2-6x-6)=0
The author's solution is quite unobvious
I used substitution.
yes I can.
Solve loga 27 + logb 4 = 5
I changed all baaes to base 5 and I got my answer write
5
I did it in about ten seconds
👍
NICE🙂
Thanks
ln x / ln 5 = 2 ln 5 / ln x
(ln x)^2 = 2 (ln 5)^2
ln x = ±√2 ln 5
x = 5^(±√2)
Yeeeee
Of course, in actual practice, solutions are NEVER this practical. Mathematicians have it easy; people's lives aren't at stake.
Have a nice day. (:
Thanks! You, too!
We're in the entertainment business 😜😁
One of these days I'll remember that square roots can also be negative...
There are two numbers whose square equals a certain positive number. That's why (or should I say y) 😁
thank you! I always forget the smaller of them.
Isn’t loga x logb=log(a+b) ???
Swap + with ×
ye?
log a + log b = log ab
log a × log b ≠ log (a + b)
a^(b+c)=a^(b)×a^(c)
No
log(ab)=log(a)+log(b)
It seemed pretty obvious to use log5basex was i/logxbase5, so I did not need your first two methods. Thanks.
Çox gözəl həll etdiniz.Bakıdan salamlar.
Çox sağ ol. Salam!