Solve for x log base 2 of 2x+ log base 2 of x=5

$\log_{2} (2 x) + \log_{2} (x) = 5$

Simplify the left side.

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

$\log_{2} (2 x \cdot x) = 5$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$\log_{2} (2 (x \cdot x)) = 5$

Multiply $x$ by $x$ .

$\log_{2} (2 x^{2}) = 5$

Rewrite $\log_{2} (2 x^{2}) = 5$ 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 $b^{y} = x$ .

$2^{5} = 2 x^{2}$

Solve for $x$ .

Rewrite the equation as $2 x^{2} = 2^{5}$ .

$2 x^{2} = 2^{5}$

Divide each term in $2 x^{2} = 2^{5}$ by $2$ and simplify.

Divide each term in $2 x^{2} = 2^{5}$ by $2$ .

$\frac{2 x^{2}}{2} = \frac{2^{5}}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x^{2}}{2} = \frac{2^{5}}{2}$

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{2^{5}}{2}$

Simplify the right side.

Cancel the common factor of $2^{5}$ and $2$ .

Factor $2$ out of $2^{5}$ .

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

Cancel the common factors.

Factor $2$ out of $2$ .

$x^{2} = \frac{2 \cdot 2^{4}}{2 (1)}$

Cancel the common factor.

$x^{2} = \frac{2 \cdot 2^{4}}{2 \cdot 1}$

Rewrite the expression.

$x^{2} = \frac{2^{4}}{1}$

Divide $2^{4}$ by $1$ .

$x^{2} = 2^{4}$

Raise $2$ to the power of $4$ .

$x^{2} = 16$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{16}$

Simplify $\pm \sqrt{16}$ .

Rewrite $16$ as $4^{2}$ .

$x = \pm \sqrt{4^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 4$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 4$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 4$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 4, - 4$

Exclude the solutions that do not make $\log_{2} (2 x) + \log_{2} (x) = 5$ true.

$x = 4$

Solve for x log base 2 of 2x+ log base 2 of x=5

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