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Alright, here's the equation:

$‎1.08^x = 1.10^{x-1}$

I know I need to use logarithms, but I can't figure how to do it. Thanks in advance!

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    I'm voting to close this question as off-topic because it is an extremely specific instance of properties of the logarithm.2015-01-30

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Don't get hung up on the fact that the bases don't match. The so-called power rule (or exponent rule) for logarithms works for any base.

$ \begin{align*} 1.08^x &= 1.10^{x-1}\\ \ln\left(1.08^x\right) &= \ln\left(1.10^{x-1}\right)\\ x\ln(1.08) &= (x-1)\ln(1.10)\\ x\ln(1.08) &= x\ln(1.10) - \ln(1.10)\\ x\ln(1.08) - x\ln(1.10) &= - \ln(1.10)\\ x(\ln(1.08) - \ln(1.10)) &= - \ln(1.10)\\ x &= \frac{-\ln(1.10)}{\ln(1.08) - \ln(1.10)} \end{align*} $

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    I h$a$ppen to have some more worked examples $f$rom a Pre-Calculus course I taught a couple years ago. I hope they are helpful to you. http://austinmohr.com/Home_files/log%20worksheet.pdf2011-06-20