How do you write 3^-2 = 1/9 in log form?

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How do you write 3^-2 = 1/9 in log form?

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Answer 1 · 2018-05-15T01:53:01

-2= ${log}_{3}$ $log_3$ (1/9)

Explanation:

A basic log transition is $x = {log}_{b} (a)$ $x=log_b(a)$ to $a = b^{x}$ $a=b^x$ .

If you plug in the variables of $a = \frac{1}{9}$ $a=1/9$ , $b = 3$ $b=3$ , and $x = - 2$ $x=-2$ , you can go from $a = b^{x}$ $a=b^x$ (or $3^{- 2} = \frac{1}{9}$ $3^-2=1/9$ if you input the variables) to $x = {log}_{b} (a)$ $x=log_b(a)$ , which is equivalent to $- 2 = {log}_{3} (\frac{1}{9})$ $-2=log_3(1/9)$ if you plug in the variables.

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How do you write 3^-2 = 1/9 in log form?

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