Physics Help Forum Solving equation for an exponent

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 Jan 20th 2009, 09:08 PM #1 Junior Member   Join Date: Jan 2009 Posts: 6 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.
 Jan 21st 2009, 11:03 AM #2 Junior Member   Join Date: Jan 2009 Posts: 6 Is this an inappropriate question for the Miscellaneous forum?
Jan 22nd 2009, 01:09 AM   #3
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 Originally Posted by Bacat 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.

 Jan 22nd 2009, 07:27 PM #4 Senior Member   Join Date: Dec 2008 Location: Las Cruces NM Posts: 256 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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