How do you solve for x in $9^{x} = \frac{1}{81}$ $9^x=1/81$ ?

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Answer 1 · 2015-12-04T18:41:34

$x = - 2$ $x=-2$

First of all, observe that $9^{2} = 81$ $9^2=81$ . Then, remember how negative powers work:

$a^{- b} = \frac{1}{a^{b}}$ $a^{-b} = 1/a^{b}$ .

So, we have that $\frac{1}{81} = \frac{1}{9^{2}} = 9^{- 2}$ $1/81 = 1/9^2 = 9^{-2}$ .

Now, we know that $9^{x} = 9^{- 2}$ $9^x=9^(-2)$ .

And so $x = - 2$ $x=-2$ .

How do you solve for x in 9x=1819^x=1/81?