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 General Physics General Physics Help Forum 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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 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.  Tags equation, exponent, solving Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post kanishka Quantum Physics 1 Oct 30th 2013 01:37 PM muon321 Quantum Physics 1 Jun 12th 2013 01:27 PM mattsco Advanced Thermodynamics 1 Sep 21st 2009 12:09 AM zpotts Special and General Relativity 2 Jul 26th 2009 07:38 PM Solid8Snake Kinematics and Dynamics 1 Apr 25th 2009 06:50 PM