If you invest $\$1500$ at $7\%$ compounded annually, how many years would it take for your investment to grow If you invest $\$1500$ at $7\%$ compounded annually, how many years would it take for your investment to grow to $3750$?
Is this right?
$$\text{term } = \frac{\log( \text{overall gain factor})}{ \log (\text{annual gain} ) }= 
\frac{ \log (3750/1500)}{ \log (1 + .07) }= \frac{\log (2.38)}{ \log (1.07) }= 12.816\text{ years}$$
 A: 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 $\frac{3750}{1500}$. 
A: 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.
A: 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.
