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

What can I do with these expression: $2^{\log _{\frac{4}{3}}n}$ and $2^{\log _{4}n}$ if I don't want to have $n$ in the exponent?

I tried nothing because I didn't have any good ideas.

Thanks.

share|cite|improve this question
up vote 2 down vote accepted

Consider that $\log_b n=\frac{\ln n}{\ln b}$ and that $b^x=e^{x \ln b}$. Combining these, $$a^{\log_b n}=e^{\ln a \ln n/\ln b}=n^{\log_b a}$$.

share|cite|improve this answer

For example, $2^{\log_4n}=2^{2\cdot\frac{1}{2}\cdot\log_4n}=((2^{2})^{\log_4n})^{\frac{1}{2}}=(4^{\log_4n})^{\frac{1}{2}}=n^{\frac{1}{2}}=\sqrt{n}.$ In this case, I am using the fact that $2^2=4$ and the property that says that $a^{\log_ab}=b.$

For the other expression, we would need to use that $2^{\log_2{\frac{4}{3}}}=\frac{4}{3}.$ Let us represent $\log_2{\frac{4}{3}}$ by $c$. Of course, $c\neq 0$. Then $2^{\log_{\frac{4}{3}}n}=2^{c\cdot\frac{1}{c}\cdot\log_{\frac{4}{3}}n}=((2^{c})^{\log_{\frac{4}{3}}n})^{\frac{1}{c}}=(\frac{4}{3})^{\log_{\frac{4}{3}}n})^{\frac{1}{c}}=n^{\frac{1}{c}}=n^{\frac{1}{\log_2{\frac{4}{3}}}}.$

share|cite|improve this answer

Your Answer

 
discard

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