Solve for x log base x of 1/9=-2

$\log_{x} (\frac{1}{9}) = - 2$

Step 1

Rewrite $\log_{x} (\frac{1}{9}) = - 2$ 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 $b^{y} = x$ .

$x^{- 2} = \frac{1}{9}$

Step 2

Solve for $x$ .

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{x^{2}} = \frac{1}{9}$

Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction.

$1 \cdot 9 = x^{2} \cdot 1$

Solve the equation for $x$ .

Rewrite the equation as $x^{2} \cdot 1 = 1 \cdot 9$ .

$x^{2} \cdot 1 = 1 \cdot 9$

Multiply $x^{2}$ by $1$ .

$x^{2} = 1 \cdot 9$

Multiply $9$ by $1$ .

$x^{2} = 9$

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

$x = \pm \sqrt{9}$

Simplify $\pm \sqrt{9}$ .

Rewrite $9$ as $3^{2}$ .

$x = \pm \sqrt{3^{2}}$

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

$x = \pm 3$

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.

$x = 3$

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

$x = - 3$

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

$x = 3, - 3$

Step 3

Exclude the solutions that do not make $\log_{x} (\frac{1}{9}) = - 2$ true.

$x = 3$

Solve for x log base x of 1/9=-2

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