Write in Exponential Form log base 7 of 1/49=-2

$\log_{7} (\frac{1}{49}) = - 2$

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, $b = 7$ , $x = \frac{1}{49}$ , and $y = - 2$ .

$b = 7$

$x = \frac{1}{49}$

$y = - 2$

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

$7^{- 2} = \frac{1}{49}$

Write in Exponential Form log base 7 of 1/49=-2

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