That's what the logarithm is for! In fact, that is how the logarithm is defined. If \(\displaystyle \frac{10000}{3}= (1.13)^t\) then, taking the logarithm of both sides, \(\displaystyle \log\left(\frac{10000}{3}\right)= log(1.13^t)= t log(1.13)\). \(\displaystyle x= \frac{\log\left(\frac{10000}{3}\right)}{log(1.13)}\).
Your calculator probably has two logarithm buttons- the "common" logarithm, base 10, and the "natural" logarithm, base e. As long as you use the same logarithm throughout a calculation, it doesn't matter which you use.
For \(\displaystyle \left(1+ \frac{r}{4}\right)^{60}= \frac{1}{5}\), since the unknown, r, is NOT in the exponent itself, you do not need a logarithm. Instead, get rid of the exponent, 60, by doing the opposite of taking the 60th power- take the 60th root or 1/60 power:
\(\displaystyle 1+ \frac{r}{4}= \left(\frac{1}{5}\right)^{1/60}\).
Now subtract 1 from both sides: \(\displaystyle \frac{r}{4}= \left(\frac{1}{5}\right)^{1/60}- 1\).
And last, of course, multiply both sides by 4: \(\displaystyle r= 4\left(\frac{1}{5}\right)^{1/60}- 4\).
That the exact solution and I, personally, would prefer it over an approximate, numerical solution. But it is not difficult to get that using a calculator. 1/5 is, of course, 0.2. The 60th root, or 1/60 power of 0.2 is, according to my calculator, 0.97353260205103889369927709540881. Multiplying by 4, that gives 3.8941304082041555747971083816352 and subtracting 4,
r= -0.10586959179584442520289161836478
