# How do i write this in logarithmic form? one ninth equals 3 Superscript negative 2

## 1/9=3^-2

Apr 5, 2018

${\log}_{3} \left(\frac{1}{9}\right) = - 2$

#### Explanation:

In exponential form

${2}^{4} = 16$

In logarithmic form

${\log}_{2} 16 = 4$

If one ninth

$\frac{1}{9} = {3}^{-} 2$

it means

$\frac{1}{9} = \frac{1}{9}$

because

${3}^{-} 2 = \frac{1}{3} ^ 2 = \frac{1}{9}$

In Exponential Form

$\frac{1}{9} = {3}^{-} 2$

it is also equal to

${3}^{-} 2 = \frac{1}{9}$

to write in Logarithmic Form, $3$ will be the base and $- 2$ is the value written as

${\log}_{3} \left(\frac{1}{9}\right) = - 2$