Question

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

Answer 1

Use logarithms.

Explanation:

Divide both sides by 100: `1.2=(1+(.032/12))^(12t)`
Simplify the inside: `1.2=(1.002bar6)^(12t)`
Convert into a logarithm: `log_"1.2"1.002bar6=12t`
Change of base formula: `(log1.002bar6)/log1.2=12t`
Divide both sides by 12: `color(blue)((log1.002bar6)/(12log1.2)=t`

Answer 2

`t= ln(6/5)/(12ln(1+.032/12))`

Explanation:

We will be using the property of logarithms that
`ln(x^n) = nln(x)`
(this is very useful for solving for variables in exponents)

`120 = 100(1+.032/12)^(12t)`

`=>120/100= 6/5 = (1+.032/12)^(12t)`

`=>ln(6/5) = ln((1+.032/12)^(12t))`

`=> ln(6/5) = 12tln(1+.032/12)` (by the property stated above)

Now, solving for `t` gives

`t = ln(6/5)/(12ln(1+.032/12))`