Evaluate log base 2 of x-6+ log base 2 of 4=3

$\log_{2} (x - 6) + \log_{2} (4) = 3$

Simplify the left side.

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

$\log_{2} ((x - 6) \cdot 4) = 3$

Apply the distributive property.

$\log_{2} (x \cdot 4 - 6 \cdot 4) = 3$

Simplify the expression.

Move $4$ to the left of $x$ .

$\log_{2} (4 \cdot x - 6 \cdot 4) = 3$

Multiply $- 6$ by $4$ .

$\log_{2} (4 x - 24) = 3$

Rewrite $\log_{2} (4 x - 24) = 3$ 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 $b^{y} = x$ .

$2^{3} = 4 x - 24$

Solve for $x$ .

Rewrite the equation as $4 x - 24 = 2^{3}$ .

$4 x - 24 = 2^{3}$

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

$4 x - 24 = 8$

Move all terms not containing $x$ to the right side of the equation.

Add $24$ to both sides of the equation.

$4 x = 8 + 24$

Add $8$ and $24$ .

$4 x = 32$

Divide each term in $4 x = 32$ by $4$ and simplify.

Divide each term in $4 x = 32$ by $4$ .

$\frac{4 x}{4} = \frac{32}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{32}{4}$

Divide $x$ by $1$ .

$x = \frac{32}{4}$

Simplify the right side.

Divide $32$ by $4$ .

$x = 8$

Evaluate log base 2 of x-6+ log base 2 of 4=3

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