How do you find n from this?

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>>8164636
>potato
yeah, about that...

log(2^200) = log(n)
n = 2^200
rite?

if not, then this
http://mathworld.wolfram.com/Logarithm.html

>>8164641
Is this the only way to deduce that? I mean I know its fairly obvious but is there any sort of way to evaluate one one side to n = . . .

>>8164649
apply 'log rules'
http://www.mathwords.com/l/logarithm_rules.htm

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>>8164653
How do you evaluate log_10 (2^200) in your head?

>>8164662
10^log_10 cancels itself out so you just get 2^200

>>8164673
What do you mean by "cancels itself out", is it in the sense that 3^5x = 3^5 which gives 5x = 5; x=1?
so log_10 (2^200) = log_10 (n) can be seen in the same way where log_10 represents the "3" and (2^200) and (n) represent the "5x" and "5"?

>>8164680
>5x = 5; x=1
that's multiplicative identity

"cancels itself out" is 'do something, then undo it'.
I'm facing north.
I turn 90deg, now I face east.
I turn -90deg, now I face north again.

>>8164683
surely mathfags will shit all over an analogy relating logarithms and rotations
sorry OP

>>8164683
I don't know whats being "done then undoing it"
log_10 (2^200) = log_10 (n), I don't see anything thats being "done and undone" it just seems like two static things that equate each other, other than n = 10^(log_10(2^200)) which I pointed out earlier but that seems more difficult than the initial question

Youre going to have to spoon feed me I don't understand what angle youre coming at

>>8164689
>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
http://www.themathpage.com/aprecalc/inverse-functions.htm
http://www.themathpage.com/aprecalc/logarithms.htm

>>8164703
>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 >>8164649 if that was the only way, then later in >>8164662
I asked if it was possible to evaluate 10^log_10(2^200) which also gives n

>>8164711
>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.

>>8164649
f(x)=log(x)
log(x) is a 1-1 function so for f(x1)=f(x2) <=> x1=x2

>>8164636
The definition of log_a(x) is a number y such that x=a^y=a^log_a(x)

>>8164683
The word you're looking for is "Inverse". They're inverse functions of each other.

>>8164662
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

>>8164636
[math]
\log_{10}(2^{200})=\log_{10}(n)\\
10^{\log_{10}(2^{200})}=10^{\log_{10}(n)}\\
2^{200}=n
[/math]

Better now?

>>8166906
Whats happening here? is that 10^log_10(2^200) or 10*log_10(2^200)

>>8166903
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]

>>8166912
10^log_10(2^200)
Powers are written in superscript.

>>8166918
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

>>8166922
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]

>>8166933
Ok, so how did you simplify 10^log_10(2^200) = 10^log_10(n)? I asked that in this >>8164711
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??

>>8166972
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