Solve for x log base 2 of 8x-x^2=4

$\log_{2} (8 x - x^{2}) = 4$

Step 1

Rewrite $\log_{2} (8 x - x^{2}) = 4$ 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 $b^{y} = x$ .

$2^{4} = 8 x - x^{2}$

Step 2

Solve for $x$ .

Rewrite the equation as $8 x - x^{2} = 2^{4}$ .

$8 x - x^{2} = 2^{4}$

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

$8 x - x^{2} = 16$

Subtract $16$ from both sides of the equation.

$8 x - x^{2} - 16 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $8 x - x^{2} - 16$ .

Reorder $8 x$ and $- x^{2}$ .

$- x^{2} + 8 x - 16 = 0$

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + 8 x - 16 = 0$

Factor $- 1$ out of $8 x$ .

$- (x^{2}) - (- 8 x) - 16 = 0$

Rewrite $- 16$ as $- 1 (16)$ .

$- (x^{2}) - (- 8 x) - 1 \cdot 16 = 0$

Factor $- 1$ out of $- (x^{2}) - (- 8 x)$ .

$- (x^{2} - 8 x) - 1 \cdot 16 = 0$

Factor $- 1$ out of $- (x^{2} - 8 x) - 1 (16)$ .

$- (x^{2} - 8 x + 16) = 0$

Factor using the perfect square rule.

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

$- (x^{2} - 8 x + 4^{2}) = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$8 x = 2 \cdot x \cdot 4$

Rewrite the polynomial.

$- (x^{2} - 2 \cdot x \cdot 4 + 4^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 4$ .

$- {(x - 4)}^{2} = 0$

Divide each term in $- {(x - 4)}^{2} = 0$ by $- 1$ and simplify.

Divide each term in $- {(x - 4)}^{2} = 0$ by $- 1$ .

$\frac{- {(x - 4)}^{2}}{- 1} = \frac{0}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{{(x - 4)}^{2}}{1} = \frac{0}{- 1}$

Divide ${(x - 4)}^{2}$ by $1$ .

${(x - 4)}^{2} = \frac{0}{- 1}$

Simplify the right side.

Divide $0$ by $- 1$ .

${(x - 4)}^{2} = 0$

Set the $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Solve for x log base 2 of 8x-x^2=4

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