Write in Exponential Form log base 9 of 1/81=-2

$\log_{9} (\frac{1}{81}) = - 2$

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, $b = 9$ , $x = \frac{1}{81}$ , and $y = - 2$ .

$b = 9$

$x = \frac{1}{81}$

$y = - 2$

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

$9^{- 2} = \frac{1}{81}$

Write in Exponential Form log base 9 of 1/81=-2

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