Solve this logarithmic equation. $2\log_6(x^{1/2}+x^{1/4}) = \log_4 x$
I don't have any good ideas how to start. Im stuck at this easy question, I don't know what is going on.
 A: If $x = t^4$ (and $t > 0$) this becomes
$$ \dfrac{2 (\log(t)+\log(t+1))}{\log 6}=\dfrac{2 \log(t^2+t)}{\log(6)} = \dfrac{4 \log(t)}{\log(4)}$$
and you can isolate $\log(t+1)$:
$$ \log(t+1) = \left(\dfrac{\log(6)}{\log{2}}-1\right) \log(t) = \dfrac{\log(3)}{\log(2)} \log(t)$$
i.e. 
$$ \dfrac{\log(t+1)}{\log(t)} = \dfrac{\log(2+1)}{\log(2)}$$
From this it's clear that $t=2$ (i.e. $x=16$) is a solution.   
If $t > 2$, we have $t + 1 < (3/2) t$, so $\log(t+1) < \log(3/2) + \log(t)$, and then $$\dfrac{\log(t+1)}{\log(t)} < \dfrac{\log(3/2)}{\log(t)} + 1 < \dfrac{\log(3/2)}{\log(2)} + 1 = \dfrac{\log(3)}{\log(2)}$$
If $1 < t < 2$, then $t + 1 > (3/2) t$, and $\log(t+1) > \log(3/2) + \log(t)$, and then
$$ \dfrac{\log(t+1)}{\log(t)} > \dfrac{\log(3/2)}{\log(t)} + 1 > \dfrac{\log(3/2)}{\log(2)} + 1 = \dfrac{\log(3)}{\log(2)}$$
And of course, the equation doesn't work for $t \le 0$ or $t=1$, while if $0 < t < 1$ then $\log(t+1)/\log(t) < 0$.  So $x=16$ is the only solution.
A: We move on to base $2$.
$$\log_6(\sqrt x+\sqrt[4] x)^2=\frac{\log_2(\sqrt x+\sqrt[4] x)^2}{\log_2 6}$$
$$\log_4 x=\frac{\log_2 x}{\log_2 4}$$ Hence
$$\frac{\log_2(\sqrt x+\sqrt[4] x)^2}{\log_2 6}= \frac{\log_2 x}{\log_2 4}\iff \frac{\log_2(\sqrt x+\sqrt[4] x)}{ \log_2 x}=\frac{\log_2 6}{4}$$
It is apparent that $x = 16$ satisfies this last equation so  then we avoid all other kinds of calculation. The answer is $\color{red}{x=16}$
