# If 2^x xx4^(x+1)=8 what is the value of x?

Mar 1, 2017

$x = \frac{1}{3}$

#### Explanation:

${2}^{x} \cdot {4}^{x + 1} = 8$

${2}^{x} \cdot {\left({2}^{2}\right)}^{x + 1} = {2}^{3}$

${2}^{x} \cdot {2}^{2 \left(x + 1\right)} = {2}^{3}$

${2}^{x + 2 x + 2} = {2}^{3}$

${2}^{\textcolor{red}{\left(3 x + 2\right)}} = {2}^{\textcolor{red}{3}}$

$3 x + 2 = 3$

$x = \frac{1}{3}$

Mar 2, 2017

High detail using first principles. Plus an alternative approach for the end.

$x = \frac{1}{3}$

#### Explanation:

Given:$\text{ } {2}^{x} \times {4}^{x + 1} = 8$

But 4 is ${2}^{2}$ so we have:

${2}^{x} \times {\left({2}^{2}\right)}^{x + 1} = 8$

${2}^{x} \times {2}^{2 \left(x + 1\right)} = 8$

${2}^{x} \times {2}^{2 x + 2} = 8$

But ${2}^{2 x + 2}$ is the same as: ${2}^{2 x} \times {2}^{2}$ giving

${2}^{x} \times {2}^{2 x} \times {2}^{2} = 8$

${2}^{3 x} \times 4 = 8$

Divide both sides by 4

${2}^{3 x} \times 4 \times \frac{1}{4} = 8 \times \frac{1}{4}$

${2}^{3 x} = {2}^{1} \text{ " rarr" } 3 x = 1 \implies x = \frac{1}{3}$
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$\textcolor{b l u e}{\text{Alternative approach}}$

You can solve directly from the above but I wish to demonstrate an alternative approach that could prove useful in different circumstances:

take loges of both sides remembering that $\log \left({2}^{3 x}\right) \to 3 x \log \left(2\right)$

$3 x \log \left(2\right) = \log \left(2\right)$

Divide both sides by $\log \left(2\right)$

$3 x = 1$

divide both sides by 3

$x = \frac{1}{3}$