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

How do I solve $2^{x-1}=3^{x+a}$? I cannot solve it and have spent an hour on it trying many different ways. Please help me! Thank you!

share|cite|improve this question
up vote 6 down vote accepted

To get the variables in a manageable place, take the logarithm of both sides of the equation (it does not matter what base you use; the power law for logarithms will let you bring the powers in front of the logarithm): $$ 2^{x-1}=3^{x+a}\iff\ln 2^{x-1}=\ln 3^{x+a}\iff (x-1)\ln 2= (x+a)\ln 3 $$

The above is valid, since logarithm functions are one-to-one, and since the first equation above has positive quantities on both sides.

Generally, if you have an equation with the variable appearing in an exponent, you can try (perhaps after a bit of algebra) taking logarithms to produce a more manageable equation as in the case above.

Finishing this problem: $$\eqalign{ &(x-1)\ln 2= (x+a)\ln 3\cr \iff& x\ln 2-\ln 2 = x\ln 3+a\ln 3\cr \iff &x\ln 2-x\ln 3= \ln 2+a\ln 3\cr \iff &x (\ln 2- \ln 3)= \ln 2+a\ln 3\cr \iff &x = {\ln 2+a\ln 3\over \ln 2- \ln 3 }\cr &= { \ln(2\cdot 3^a)\over \ln(2/3)}. } $$

share|cite|improve this answer
+1. In fact ${ \log(2\cdot 3^a)\over \log(2/3)}$ to any base – Henry Dec 24 '11 at 12:52
Thank you so much David! ^^ – tina nyaa Dec 25 '11 at 1:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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