Evaluate log base x of 8=-1/2

$\log_{x} (8) = - \frac{1}{2}$

Step 1

Rewrite $\log_{x} (8) = - \frac{1}{2}$ 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 $b^{y} = x$ .

$x^{- \frac{1}{2}} = 8$

Step 2

Solve for $x$ .

Raise each side of the equation to the power of $- 2$ to eliminate the fractional exponent on the left side.

${(x^{- \frac{1}{2}})}^{- 2} = 8^{- 2}$

Simplify the exponent.

Simplify the left side.

Simplify ${(x^{- \frac{1}{2}})}^{- 2}$ .

Multiply the exponents in ${(x^{- \frac{1}{2}})}^{- 2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x^{- \frac{1}{2} \cdot - 2} = 8^{- 2}$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$x^{\frac{- 1}{2} \cdot - 2} = 8^{- 2}$

Factor $2$ out of $- 2$ .

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

Cancel the common factor.

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

Rewrite the expression.

$x^{- 1 \cdot - 1} = 8^{- 2}$

Multiply $- 1$ by $- 1$ .

$x^{1} = 8^{- 2}$

Simplify.

$x = 8^{- 2}$

Simplify the right side.

Simplify $8^{- 2}$ .

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

$x = \frac{1}{8^{2}}$

Raise $8$ to the power of $2$ .

$x = \frac{1}{64}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1}{64}$

Decimal Form:

$x = 0.015625$

Evaluate log base x of 8=-1/2

Related Questions

Support LCMGCF.com by disabling your adblocker.