# Solving a variable for an exponential function

So I have the following problem:

A person puts 1,100 dollars in an account that he expects will make 5% interest in periods of three months. He then realizes a year later he was wrong about the interest rate and has $50 less than expected. What is the actual interest rate? So I was able to get an equation out of that: $$A_1 = 1100(1 + \frac{0.05}{4})^{4t}$$ $$A_1 - 50 = 1100(1 + \frac{r}{4})^{4t}$$ $$1100(1 + \frac{0.05}{4})^{4t} - 50 = 1100(1 + \frac{r}{4})^{4t}$$ Now there is where I get stuck on how to simplify the equation for r? • wow what is going on... you denote the interest by$r$? if so what is$t$? and how you got that$4$? – Yanko Aug 7 '17 at 20:41 • Note: the second equation should be$A_1 + 50$. Are you allowed to use a calculator/computational device for this? – platty Aug 7 '17 at 20:41 • I am allowed to use a scientific calculator @platty – Pablo Aug 7 '17 at 20:41 • @platty do you understand the first equation? why he divides by 4? and what's$t$? – Yanko Aug 7 '17 at 20:42 • There is not enough information. When does he learn that he has made$\$50$ less than expected? – Doug M Aug 7 '17 at 20:43

You are looking for $r$ which makes $$f(r,t)=1100\left(1 + \frac{0.05}{4}\right)^{4t} - 50 - 1100\left(1 + \frac{r}{4}\right)^{4t}=0$$ The solution depends on $t$.
There is no explicit solution of this equation except if $t=1$ (just as Doug M commented) and numerical methods are required. This would be "simple" since the solution should be "close" to $r=0.05$ if $t$ is sufficiently large (as it uses to be in investment problems).
Assuming it (and making all numbers rational), we can use Taylor series limited to first order and get as a very first approximation $$r=\frac 1 {20}-\frac{2^{16 t-5} \left(\frac{81}{5}\right)^{1-4 t}}{11 t}\tag 1$$ Better would be the simplest Padé approximant which would give $$r=\frac 1 {20}-\frac{81\ 2^{16 t-1} 5^{4 t-1}}{\left(176\ 81^{4 t}-2^{16 t+2} 5^{4 t}\right) t+2^{16 t} 5^{4 t}}\tag 2$$ But the rigorous way would be to use a numerical method (such as Newton) starting using $r_0=0.05$.
I give you below a table of results $$\left( \begin{array}{cccc} t & (1) & (2) & \text{exact} \\ 1 & 0.0062083 & 0.0054863 & 0.0054795 \\ 2 & 0.0291655 & 0.0287835 & 0.0287805 \\ 3 & 0.0367837 & 0.0365421 & 0.0365404 \\ 4 & 0.0405683 & 0.0404006 & 0.0403994 \\ 5 & 0.0428204 & 0.0426974 & 0.0426966 \\ 6 & 0.044307 & 0.0442135 & 0.0442129 \\ 7 & 0.0453568 & 0.0452838 & 0.0452834 \\ 8 & 0.0461342 & 0.0460761 & 0.0460758 \\ 9 & 0.0467303 & 0.0466834 & 0.0466832 \\ 10 & 0.0471999 & 0.0471616 & 0.0471615 \\ 11 & 0.0475779 & 0.0475463 & 0.0475462 \\ 12 & 0.0478873 & 0.0478611 & 0.047861 \\ 13 & 0.0481444 & 0.0481225 & 0.0481224 \\ 14 & 0.0483605 & 0.048342 & 0.0483419 \\ 15 & 0.0485439 & 0.0485283 & 0.0485283 \end{array} \right)$$