Solve for x log base 2 of x^2-9=4

$\log_{2} (x^{2} - 9) = 4$

Step 1

Rewrite $\log_{2} (x^{2} - 9) = 4$ 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 $b^{y} = x$ .

$2^{4} = x^{2} - 9$

Step 2

Solve for $x$ .

Rewrite the equation as $x^{2} - 9 = 2^{4}$ .

$x^{2} - 9 = 2^{4}$

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

$x^{2} - 9 = 16$

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

Add $9$ to both sides of the equation.

$x^{2} = 16 + 9$

Add $16$ and $9$ .

$x^{2} = 25$

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

$x = \pm \sqrt{25}$

Simplify $\pm \sqrt{25}$ .

Rewrite $25$ as $5^{2}$ .

$x = \pm \sqrt{5^{2}}$

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

$x = \pm 5$

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.

$x = 5$

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

$x = - 5$

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

$x = 5, - 5$

Solve for x log base 2 of x^2-9=4

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