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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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closed as off-topic by user2345215, Lord_Farin, dustin, Live Forever, Eric Stucky Jan 31 '15 at 2:15

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Begin by taking the log of both sides. Then look at the property of logs to see if you can reduce either of the sides to something nicer! – james Jun 20 '11 at 1:26
Your right-hand side is ambiguous. Is the exponent $x-1$, or is the exponent $x$? I.e., is the equation $1.08^x=1.10^{x-1}$ or $1.08^x=1.10^x-1$? – Jonas Meyer Jun 20 '11 at 1:26
The left one, sorry Jonas. – user576488 Jun 20 '11 at 1:28
I'm voting to close this question as off-topic because it is an extremely specific instance of properties of the logarithm. – Lord_Farin Jan 30 '15 at 23:14
up vote 4 down vote accepted

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 happen to have some more worked examples from a Pre-Calculus course I taught a couple years ago. I hope they are helpful to you. – Austin Mohr Jun 20 '11 at 2:33

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