How do you find n from this?

>log_10 (2^200) = log_10 (n), nothing done and undone
correct, in code monkey terms merely note that log_10(var) is a function and that the arguments passed to the function are equal.

the exponential function is the inverse of the log function
themathpage.com/aprecalc/inverse-functions.htm
themathpage.com/aprecalc/logarithms.htm

>correct, in code monkey terms merely note that log_10(var) is a function and that the arguments passed to the function are equal.

yes I understand that is the way to realize what n=, but I mentioned in if that was the only way, then later in I asked if it was possible to evaluate 10^log_10(2^200) which also gives n

>if it was possible to evaluate 10^log_10(2^200)
Lel, yes and doing so relies on recognition of a function and its inverse.

This post marks the end of my serious replies to the thread.
Best of luck realizing whatever goal you have.

f(x)=log(x)
log(x) is a 1-1 function so for f(x1)=f(x2) x1=x2

The definition of log_a(x) is a number y such that x=a^y=a^log_a(x)

The word you're looking for is "Inverse". They're inverse functions of each other.

log(2^200) gives you the number you have to raise 10 to to gain 2^200. So obviously if you then actually raise 10 to that number you will gain 2^200

I'm retarded

How do I work x out of this

10 (-95/61.5) = 4/x

I thought I had to divide everything by 4 but it's not giving me the right answer (Trying to do the Nernst equation)

The correct answer is 140, but I don't know how to get it. I emailed a lecturer about it and she said

>"You need to take 1 over the calculated value to find Kout"
wtf does this mean?

I've always been retarded trying to do basic maths.

>The Nernst equation is Eq = 61.5 log (Kout/Kin)

In this situation

Eq = -95
Kout = 4
Kin = X (Answer is 140)

I post here because I'm out of ideas

[math]
\log_{10}(2^{200})=\log_{10}(n)\\
10^{\log_{10}(2^{200})}=10^{\log_{10}(n)}\\
2^{200}=n
[/math]

Better now?

Whats happening here? is that 10^log_10(2^200) or 10*log_10(2^200)

the solution to that equation is not 140
[math]
10\left (\frac{-95}{61.5} \right ) = \frac{4}{x}\\
10x\left (\frac{-95}{61.5} \right ) = 4\\
-x\frac{950}{61.5}=4\\
x=-\frac{123}{475}
[/math]

however,
[math]
-95 = 61.5 \log_{10}\left (\frac{4}{x} \right )\\
\frac{-95}{61.5} = \log_{10}\left (\frac{4}{x} \right )\\
10^{\frac{-95}{61.5}}=10^{\log_{10}\left (\frac{4}{x} \right )}\\
10^{\frac{-95}{61.5}}=\frac{4}{x}\\
x=\frac{4}{10^{\frac{-95}{61.5}}}\\
x= 4*10^{\frac{95}{65.5}}\\
x=140.20885
[/math]

10^log_10(2^200)
Powers are written in superscript.

The font is kinda funny and looks semi super scripted, just wanted to check. But can you explain to me why you multiplied both sides by 10? how does 10*log_10(n) turn into n? how do you know that 10^(10*log_10(2^200)) = 10

it is not a multiplication
[math]
l=\log_{n}(a)\\
n^{l}=a
[/math]
Logarithms answer the question: How much do I have to raise some base to be this number. Therefore
[math]
n^{\log_{n}(a)}=a\\
[/math]

Ok, so how did you simplify 10^log_10(2^200) = 10^log_10(n)? I asked that in this but no one really said anything, how did you do this in your head?

when you have 10^log_10(x) = 10^log_10(y) x and y must be the same... so it just gives you the original equation back in a never ending loop??

what? 10^log_10(2^200) = 10^log_10(n) becaus you said so.

If
log_10(2^200) = log_10(n)
then
10^log_10(2^200) = 10^log_10(n)
and then, using the logarithm definition,
2^200 = n