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If

$\log_2(x)=7~\text{find}~ \log_8(x)=y$

$\begin{array}{|lrcll|} \hline (1) & x &=& 2^7 \qquad \Big(\log_2(x)=7\Big) \\\\ (2) & x &=& 8^y \qquad \Big(\log_8(x)=y\Big) \\ & x &=& 2^{3y}\\ \hline & x = 2^7 &=& 2^{3y}\\ & 2^7 &=& 2^{3y} \quad & | \quad \text{compare exponent}\\ & 7 &=& 3 y \\ & \mathbf{y} &=& \dfrac{7}{3} \\ & \mathbf{\log_8(x)} &=& \dfrac{7}{3} \\ \hline \end{array}$