Solve $8^{x+1} = 32 \cdot\sqrt2$ without $\log$ I need help solving the equation $8^{x+1} = 32 \cdot\sqrt2$. The obvious answer is to use log, but that is reserved for the next section. The example given for this section of questions is:
$4^x = 8$
$(2^2)^x = 2^3$
$2x = 3$
$x = \dfrac{3}{2}$
The example looks obvious and easy to solve, but I do not understand where you'd use this for my question.

The idea is that everything in sight is a power of $2$.
$$
\begin{eqnarray*}
  8^{x+1} &=& 32 \sqrt{2} \\
  (2^3)^{x+1} &=& 2^5\cdot 2^{1/2} \\
  2^{3(x+1)} &=& 2^{5 + 1/2}\\
\end{eqnarray*}
$$
Since both sides are expressed as a power of the same base, we can set the exponents equal:
$$
\begin{eqnarray*}
  3(x+1) &=& \frac{11}{2} \\
  x+1 &=& \frac{11}{6}\\
  x &=& \frac{11}{6} - 1 \\
  &=& \frac{5}{6}
\end{eqnarray*}
$$
