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If $\log_x(4x^{\log_5x}+5)=2\log_5x$, then find the value of $x$

I could proceed thus:

$$\log_x(4x^{\log_5x}+5)=2\log_5x$$ $$\therefore \log_5(4x^{\log_5x}+5)=2(\log_5x)^2$$

Now, I don't know how to simplify this.

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HINT:

Let $x=5^y\implies \log_5x=y$

Using $\displaystyle\log_ab=\frac{\log b}{\log a}\implies \log_x(4x^{\log_5x}+5)=\frac{\log(4x^{\log_5x}+5)}{\log x}$

$$\implies\frac{\log (4\cdot( 5^y)^y+5)}{\log (5^y)}=2y$$

Using $ m\log b=\log b^m $ ( if $ \log b$ exists) in the Right Hand Side,

$$\implies\log (4\cdot( 5^y)^y+5)=2y\log (5^y)=\log(5^y)^{2y}$$

Using $\displaystyle\log a=\log b\iff a=b$ for positive real $a,b$

$$\implies 4\cdot5^{y^2}+5 =(5)^{2y^2}=(5^{y^2})^2$$

Setting $\displaystyle5^{(y^2)}=z,$ we have $\displaystyle z^2-4z-5=0$

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  • $\begingroup$ It should be $z^2-4z-5=0$. Solving, we get $x=5,\frac15$ $\endgroup$ – Tejas Dec 25 '13 at 4:09
  • $\begingroup$ @TejasAdsul, thanks for your observation. But the roots of $z^2-4z-5=0$ are $5,-1$ $\endgroup$ – lab bhattacharjee Dec 25 '13 at 4:15
  • $\begingroup$ Yes. But they are the solutions of $z$. From $z$, we will find $y$, which will be $1,-1$. And then $x$. $\endgroup$ – Tejas Dec 25 '13 at 4:16
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    $\begingroup$ @TejasAdsul, Right. If $z=-1, 5^{(y^2)}=-1$ which leads to non-real solutions $\endgroup$ – lab bhattacharjee Dec 25 '13 at 4:20

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