Solve for x log base 2 of 2x+ log base 2 of 2=3

$\log_{2} (2 x) + \log_{2} (2) = 3$

Step 1

Simplify the left side.

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log_{2} (2 x \cdot 2) = 3$

Multiply $2$ by $2$ .

$\log_{2} (4 x) = 3$

Step 2

Rewrite $\log_{2} (4 x) = 3$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$2^{3} = 4 x$

Step 3

Solve for $x$ .

Rewrite the equation as $4 x = 2^{3}$ .

$4 x = 2^{3}$

Divide each term in $4 x = 2^{3}$ by $4$ and simplify.

Divide each term in $4 x = 2^{3}$ by $4$ .

$\frac{4 x}{4} = \frac{2^{3}}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{2^{3}}{4}$

Divide $x$ by $1$ .

$x = \frac{2^{3}}{4}$

Simplify the right side.

Raise $2$ to the power of $3$ .

$x = \frac{8}{4}$

Divide $8$ by $4$ .

$x = 2$

Solve for x log base 2 of 2x+ log base 2 of 2=3

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