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.