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Old Jan 20th 2009, 09:08 PM   #1
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Solving equation for an exponent

This is actually a math problem, but it's useful in chemical physics. I can't remember how to solve an equation of the following form:

$\displaystyle ln(a^x) = ln(b^x)$

I want to solve this for x where a and b are both real numbers. I know I can manipulate the logarithm to write:

$\displaystyle xln(a) = xln(b)$

But this just seems to cancel out the x's. If you can show me how to solve this type of equation for x, I would very much appreciate it.

Thank you for your time.
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Old Jan 21st 2009, 11:03 AM   #2
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Is this an inappropriate question for the Miscellaneous forum?
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Old Jan 22nd 2009, 01:09 AM   #3
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Originally Posted by Bacat View Post
This is actually a math problem, but it's useful in chemical physics. I can't remember how to solve an equation of the following form:

$\displaystyle ln(a^x) = ln(b^x)$

I want to solve this for x where a and b are both real numbers. I know I can manipulate the logarithm to write:

$\displaystyle xln(a) = xln(b)$

But this just seems to cancel out the x's. If you can show me how to solve this type of equation for x, I would very much appreciate it.

Thank you for your time.
$\displaystyle ln(a^x) - ln(b^x)=0$
$\displaystyle ln(\frac{a^x}{b^x})=0$
$\displaystyle xln(\frac{a}{b})=0$
Suppose $\displaystyle \frac{a}{b}$ is positive, then x=0?
I don't know if i was doing the right thing as my maths is quite poor.
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Old Jan 22nd 2009, 07:27 PM   #4
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Going from $\displaystyle ln(a^x) = ln(b^x) $
to $\displaystyle x ln(a) = x ln(b) $
assumes $\displaystyle a > 0 $ and $\displaystyle b > 0 $

e.g. $\displaystyle ln((-2)^2) = ln((2)^2) $
but
$\displaystyle 2 ln(-2) = 2 ln(2) $ makes no sense.)

If we assume the above, then the equation has no solution for $\displaystyle x $ unless $\displaystyle a = b $.

If $\displaystyle a = b $ then any real number x is a solution.
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