If Joe has $\$1500$ to invest at $12\%$ per year compounded monthly, how long will it be before he has $\$2100?$

Question

Here is my work so far: $A=P(1+r/n)^{nt}$

$2100=1500(1+0.12/12)^{12t}$

$2100/1500=(1.01)^{12t}$

Here is the part that I am stuck on. Would I take the natural log of both sides? And in general when solving for problems similar to the one that I have given, when would I take the natural log of both sides

Answer 1

You can use logs when you have positive quantities, which in this type of problems always happens. So :

$$A=P(1+\frac{r}{n})^{nt} \iff$$ $$ \log{\frac{A}{P}}= nt \log{(1+\frac{r}{n} )} \iff $$ $$ t = \frac{\log{A}-\log{P}}{n\log{(1+\frac{r}{n})}}$$

This is the general formula for $t$.

Answer 2

$$ \frac{2100}{1500} = (1.01)^{12t} $$

$$ \text{ln}(\frac{2100}{1500}) = \text{ln}((1.01)^{12t}) $$

$$ \text{ln}(\frac{2100}{1500}) = 12t \times \text{ln}(1.01) $$

$$ \frac{\text{ln}(\frac{2100}{1500})}{\text{ln}(1.01)} = 12t $$

$$ t = \frac{1}{12} \times \frac{\text{ln}(\frac{2100}{1500})}{\text{ln}(1.01)}. $$