Arithmetic progression - Logarithm

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

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

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

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

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

$$2\log_mx=\log_nx+\log_kx$$

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

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

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

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

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

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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