Solve $8^{x+1} = 32 \cdot\sqrt2$ without $\log$

Question

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: \begin{align*} 4^x & = 8\\ (2^2)^x & = 2^3\\ 2x & = 3\\ x & = \dfrac{3}{2} \end{align*}

The example looks obvious and easy to solve, but I do not understand where you'd use this for my question.

Answer 1

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: $$ \begin{eqnarray*} 3(x+1) &=& \frac{11}{2} \\ x+1 &=& \frac{11}{6}\\ x &=& \frac{11}{6} - 1 \\ &=& \frac{5}{6} \end{eqnarray*} $$

Answer 2

$$ 8^{x+1} = 32 \sqrt{2} $$ $$ 8\times 8^x = 32 \sqrt{2} $$ $$ 8^x =2^{3x}= 4\sqrt{2}=2^2\times 2^{1/2}= 2^{2,5} $$

$$ 3x=2,5 \to x=\frac{5}{6} $$