# How do you solve 120=100(1+(.032/12))^(12t)?

Nov 14, 2015

Use logarithms.

#### Explanation:

Divide both sides by 100: $1.2 = {\left(1 + \left(\frac{.032}{12}\right)\right)}^{12 t}$
Simplify the inside: $1.2 = {\left(1.002 \overline{6}\right)}^{12 t}$
Convert into a logarithm: ${\log}_{\text{1.2}} 1.002 \overline{6} = 12 t$
Change of base formula: $\frac{\log 1.002 \overline{6}}{\log} 1.2 = 12 t$
Divide both sides by 12: color(blue)((log1.002bar6)/(12log1.2)=t

Nov 14, 2015

$t = \ln \frac{\frac{6}{5}}{12 \ln \left(1 + \frac{.032}{12}\right)}$

#### Explanation:

We will be using the property of logarithms that
$\ln \left({x}^{n}\right) = n \ln \left(x\right)$
(this is very useful for solving for variables in exponents)

$120 = 100 {\left(1 + \frac{.032}{12}\right)}^{12 t}$

$\implies \frac{120}{100} = \frac{6}{5} = {\left(1 + \frac{.032}{12}\right)}^{12 t}$

$\implies \ln \left(\frac{6}{5}\right) = \ln \left({\left(1 + \frac{.032}{12}\right)}^{12 t}\right)$

$\implies \ln \left(\frac{6}{5}\right) = 12 t \ln \left(1 + \frac{.032}{12}\right)$ (by the property stated above)

Now, solving for $t$ gives

$t = \ln \frac{\frac{6}{5}}{12 \ln \left(1 + \frac{.032}{12}\right)}$