# 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.

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

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damn you were faster... –  draks ... Nov 22 '11 at 21:33
If it makes you feel better, I read through both of them :) –  erimar77 Nov 22 '11 at 22:43

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

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

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