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If $\log_{k}x+\log_{n}x=2\log_{m}x$ than $n^2=(kn)^{\log_{k}m}$, for all $x,m,n,k>0$ and $x,m,n,k \neq 1$

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closed as off-topic by Your Ad Here, user127.0.0.1, TZakrevskiy, Sami Ben Romdhane, egreg Feb 12 '14 at 23:28

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For logarithms the following equation holds: $$\log_kx=\frac{\ln x}{\ln k}$$ Now the first equations is after dividing by $\ln x$ $$\frac{1}{\ln k}+\frac{1}{\ln n}=\frac 2{\ln{m}}\Rightarrow \frac{\ln{m}}{\ln k}=2-\frac{\ln{m}}{\ln n}$$ In the equation we want to prove we have on the RHS $$n^{\ln m/ \ln k}\cdot k^{\log_k m}=n^{2-\ln m/ \ln n}\cdot m=\frac{n^2}{n^{\log_n m}}\cdot m = n^2$$ what equals the LHS.

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