Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user2345215, dustin, ᴡᴏʀᴅs
If this question can be reworded to fit the rules in the help center, please edit the question.

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.

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

share|cite|improve this answer
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

Not the answer you're looking for? Browse other questions tagged or ask your own question.