# Solve $\log_x (4x^{\log_5 x} + 5) = 2log_5 x$

My attempt

$$\log_5(4x^{log_5x}+5)=2(log_5x)^2$$ Using $$\log_ab=\frac{\log_cb}{\log_ca}$$

$$4x^{log_5x}+5=5^{2(log_5x)^2}$$

Let $$f=x^{log_5x}\Rightarrow \log_5 f = (log_5x)^2 \Rightarrow f= 5^{(\log_5 x)^2}$$

We arrive at,

$$4 *5^{(\log_5 x)^2}+5=5^{2(log_5x)^2};$$ Let $$5^{(\log_5 x)^2} = z$$

$$z^2-4z-5=0$$ It gives;

$$z=-1, 5$$

$$z=5$$ gives $$x=5, \frac{1}{5}.$$ [Real solution]

$$z=-1$$ gives $$5^{\sqrt{\log_5 (-1)}} = -0.0941629 - 4.90284 I$$ [Complex solution]

Ques: I am suspecting that this complex solution should not be there, Could you please help me ?

It is so simple, check the domain and range of $$5^{(\log_5 x)^2}$$
We know that $$(\log_5 x)^2\ge0$$ So, we get $$5^{(\log_5 x)^2}\ge1$$ or $$z\ge 1$$
So, $$z=5$$ is the only solution.