Write in Exponential Form log base 8 of 1/8=-1

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

For logarithmic equations, $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ such that $x > 0$ , $b > 0$ , and $b \neq 1$ . In this case, $b = 8$ , $x = \frac{1}{8}$ , and $y = - 1$ .

$b = 8$

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

$y = - 1$

Substitute the values of $b$ , $x$ , and $y$ into the equation $b^{y} = x$ .

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

Write in Exponential Form log base 8 of 1/8=-1

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