Write in Exponential Form -3 = log base 9 of 1/729

$- 3 = \log_{9} (\frac{1}{729})$

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, $b = 9$ , $x = \frac{1}{729}$ , and $y = - 3$ .

$b = 9$

$x = \frac{1}{729}$

$y = - 3$

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

$9^{- 3} = \frac{1}{729}$

Write in Exponential Form -3 = log base 9 of 1/729

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