# How do you solve 2.1^(t-5)=9.32?

Dec 10, 2016

I have taken you to a point where you can complete it

#### Explanation:

Using the principle that $\log \left({a}^{b}\right) \to b \log \left(a\right)$

Take logs of both sides

$\left(t - 5\right) \log \left(2.1\right) = \log \left(9.23\right)$

$t - 5 = \frac{\log \left(9.23\right)}{\log \left(2.1\right)}$

$t = \frac{\log \left(9.23\right)}{\log \left(2.1\right)} + 5$