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.
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.
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 .....
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.
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)
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 am not good at math, so I watch these videos to learn. Hopefully I get better.
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
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^(±√2)
I did in mind in 10 seconds!!!
Wow! 😁🤩
@@SyberMath :D
There is a small error at the 2nd method
X=5^(sqrt(2)) or 5^(-sqrt(2))
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 😜😁
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
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 .....
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=
I changed all baaes to base 5 and I got my answer write
I couldn't understand why 2log5's 2 become root 2 in next line's square...why not instead of 2???
that was great! 😍 i think i did a 4th method but pretty close to method 1. Awesome video
Awesome! Thank you!
Nice!
I made up this equation but I don't know how to solve it
Equation:
logx(5x)=2x/5
Answer: 5
i naturally went straight to the 2nd method. i'm a "change of base" kinda guy.
I used change of base with 5 as the base. Led to the same result, 5^(+/-√2)
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.
Thank you!!!
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
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)
ln x / ln 5 = 2 ln 5 / ln x
(ln x)^2 = 2 (ln 5)^2
ln x = ±√2 ln 5
x = 5^(±√2)
x = 5^(sqrt)
it's 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
Solve loga 27 + logb 4 = 5
5
Hello teacher 🙏
Hello! 🤩
I used substitution.
yes I can.
Yeeeee
Confusing!
Why? or what part?
I did it in about ten seconds
👍
X=5^(sqrt2),x=5^(-sqrt2)
y?
NICE🙂
Thanks
Not any substitution needed,nor any reciprocal rule,flipping etc. The second method is simple and perfect as well
Good to hear!
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!