Solve for x log base 2 of x-7+ log base 2 of x=3

$\log_{2} (x - 7) + \log_{2} (x) = 3$

Simplify the left side.

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

$\log_{2} ((x - 7) x) = 3$

Apply the distributive property.

$\log_{2} (x \cdot x - 7 x) = 3$

Multiply $x$ by $x$ .

$\log_{2} (x^{2} - 7 x) = 3$

Rewrite $\log_{2} (x^{2} - 7 x) = 3$ 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 $b^{y} = x$ .

$2^{3} = x^{2} - 7 x$

Solve for $x$ .

Rewrite the equation as $x^{2} - 7 x = 2^{3}$ .

$x^{2} - 7 x = 2^{3}$

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

$x^{2} - 7 x = 8$

Subtract $8$ from both sides of the equation.

$x^{2} - 7 x - 8 = 0$

Factor $x^{2} - 7 x - 8$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 8$ and whose sum is $- 7$ .

$- 8, 1$

Write the factored form using these integers.

$(x - 8) (x + 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 8 = 0$

$x + 1 = 0$

Set $x - 8$ equal to $0$ and solve for $x$ .

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

Add $8$ to both sides of the equation.

$x = 8$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $(x - 8) (x + 1) = 0$ true.

$x = 8, - 1$

Exclude the solutions that do not make $\log_{2} (x - 7) + \log_{2} (x) = 3$ true.

$x = 8$

Solve for x log base 2 of x-7+ log base 2 of x=3

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