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

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

Move all the terms containing a logarithm to the left side of the equation.

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

Simplify the left side.

Simplify $2 \log_{2} (x) - \log_{2} (2 x)$ .

Simplify $2 \log_{2} (x)$ by moving $2$ inside the logarithm.

$\log_{2} (x^{2}) - \log_{2} (2 x) = 3$

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{2} (\frac{x^{2}}{2 x}) = 3$

Reduce the expression $\frac{x^{2}}{2 x}$ by cancelling the common factors.

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

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

Factor $x$ out of $2 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Rewrite $\frac{x}{2}$ as $x \cdot 2^{- 1}$ .

$\log_{2} (x \cdot 2^{- 1}) = 3$

Rewrite $\log_{2} (x \cdot 2^{- 1})$ as $\log_{2} (x) + \log_{2} (2^{- 1})$ .

$\log_{2} (x) + \log_{2} (2^{- 1}) = 3$

Use logarithm rules to move $- 1$ out of the exponent.

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

Logarithm base $2$ of $2$ is $1$ .

$\log_{2} (x) - 1 \cdot 1 = 3$

Multiply $- 1$ by $1$ .

$\log_{2} (x) - 1 = 3$

Move all terms not containing $\log_{2} (x)$ to the right side of the equation.

Add $1$ to both sides of the equation.

$\log_{2} (x) = 3 + 1$

Add $3$ and $1$ .

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

Rewrite $\log_{2} (x) = 4$ 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 $b^{y} = x$ .

$2^{4} = x$

Solve for $x$ .

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

$x = 2^{4}$

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

$x = 16$

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

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