$x={}$ $8$

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The given equation is: $$\log_{2}(x)+\log_{8}(x)=4$$ The first step is to equalize the bases. It doesn't matter what base you take. Here we set the bases equal to $2$. So: $$\log_{2}(x)+\frac{\log_{2}(x)}{\log_{2}(8)}=4$$ The left-hand side can be simplified: $$\begin{aligned}\log_{2}(x)+\frac{\log_{2}(x)}{\log_{2}(2^{3})}&=\log_{2}(x)+\frac{\log_{2}(x)}{3}\\[0.25cm] &=\log_{2}(x)\bigg(1+\frac{1}{3}\bigg)=\frac{4}{3}\log_{2}(x)\\[0.25cm]\end{aligned}$$ So $$\frac{4}{3}\log_{2}(x)=4$$ Thus: $$\log_{2}(x)=3$$ On the basis of the definition of the logarithm with base 2 we get: $$\begin{aligned}x&=2^{3}\\[0.25cm] &=8\end{aligned}$$