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

If $\,\log_kx ,\, \log_mx,\,\log_nx\,$ are in A.P then prove that $n^2=(kn)^{\log_kn}$

$2\log_mx = \log_kx+\log_nx$ $\frac{\log_kx}{\log_km}$

$=\log_kx+\frac{\log_kx}{\log_kn}$ =$\frac{2}{\log_kx}=1+\frac{1}{\log_kn}$

=$\log_km[\log_k(nk)] =\log_kn^2$

Please guide further...

share|cite|improve this question

There is a typo in the Right Hand Side. It will be $\log_km$ instead of $\log_kn$


$$2\frac{\log x}{\log m}= \frac{\log x}{\log n}+\frac{\log x}{\log k}\text { as }\log_ab=\frac{\log b}{\log a}\text{ with any base }>0,\ne1$$

$$\frac2{\log m}= \frac1{\log n}+\frac1{\log k}=\frac{\log k+\log n}{\log n\log k}=\frac{\log kn}{\log n\log k}$$ as $\log x=0\iff x=1$ would make the given condition an identity, hence $\log x\ne0$

$$\text{ So, }2\log n=\frac{\log m}{\log k}\log kn$$

$$\implies\log n^2=\log_k m\log kn=\log (kn)^{\log_k m}\text { as } r\log a=\log a^r$$

$$\implies n^2= (kn)^{\log_k m}$$

share|cite|improve this answer
thanks a lot .. that was really a very good explanation....thanks a ton once again....GOD bless you... – Sachin Sharmaa Mar 4 '13 at 14:19
@SachinSharmaa, my pleasure. Thanks for the last statement. – lab bhattacharjee Mar 4 '13 at 14:20
@SachinSharmaa, observe that the proposition to be proved does not contain $x,$ so our first goal would be to eliminate $x$ – lab bhattacharjee Mar 4 '13 at 14:22

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.