# How do you solve 3^(n-2)=27?

Oct 13, 2016

$n = 5$

#### Explanation:

As ${3}^{n - 2} = 27$

converting it into logarithmic form, we have

${\log}_{3} 27 = n - 2$

But ${\log}_{3} 27 = {\log}_{3} {3}^{3} = 3 {\log}_{3} 3 = 3 \times 1 = 3$

Hence $n - 2 = 3$

and $n = 3 + 2 = 5$