If you invest $\$1500$ at $7\%$ compounded annually, how many years would it take for your investment to grow

Question

If you invest $\$1500$ at $7\%$ compounded annually, how many years would it take for your investment to grow to $3750$?

Is this right?

$$\text{term } = \frac{\log( \text{overall gain factor})}{ \log (\text{annual gain} ) }= \frac{ \log (3750/1500)}{ \log (1 + .07) }= \frac{\log (2.38)}{ \log (1.07) }= 12.816\text{ years}$$

Answer 1

When one does mathematics, it can be useful to go back to basic principles.

With interest rate of $0.07$, that is, $7\%$, compounded annually, in $n$ years $A$ dollars grow to $$A(1.07)^n$$ dollars. In our case, $1500$ grew to $3750$ in an unknown number $n$ of years, so $$3750=1500(1.07)^n.$$ It follows that $$(1.07)^n=\frac{3750}{1500}=2.5.$$ Take the logarithm of both sides, using your favourite base. We get $$n\log(1.07)=\log(2.5),$$ and therefore $$n=\frac{\log(2.5)}{\log(1.07)}.$$

Remark: Your procedure was correct. There was a little numerical slip in calculating $\frac{3750}{1500}$.

Answer 2

This is good, subject to revising your typo. You have after $n$ periods that your balance is $$1500* 1.07^n.$$ Now set $$1500*1.07^n = 3750.$$ Begin by dividing to get $$1.07^n = 2.5 $$ so $$n = {\log(2.5)\over \log(1.07)} = 13.54.$$ Your answer is off a bit because you transposed two digits.

Answer 3

If by chance you didn't want to use logarithm's you could make a table and write the formula out for the desired years:

$a(1+r)^n$

where $a$ is the initial amount, $r$ is the rate and $n$ is the number of years.

For year 12: $1500(1+0.07)^{12}$ = 3378.29

For year 13: $1500(1+0.07)^{13}$ = 3614.77

For year 14: $1500(1+0.07)^{14}$ = 3867.29

So if the desired final amount was 3750, it would take 14 years assuming we didn't want half years.

Assuming we did, simply play guess and check until we get a closer result.

$1500(1+0.07)^{13.5429}$ = 3750.01

And so on.