Ways to solve compounding interest questions?

Question

Ed invests $\$15,000$ into an account that pays an annual rate of $3.5\%$ interest compounded k times per year where $k\in \mathbb{Z}$. At the end of two years the investment is worth $\$16,087$. Find the smallest value of $k$.

For this question, I solved it using my TI-84's graphing calculator, and got $k=32$ by finding the intersection. My two equations are: $$ Y_1 = 16087\\ Y_2 = 15000(1+\dfrac{3.5}{100x})^{2x} $$

However, there is also a "TVM solver" function on the calculator, and I am wondering if there is a way to rearrange my equation to solve for $x$ using the TVM solver. I attempted to solve it by setting: $$ N=2\\ I=3.5\\ PV=15000\\ FV=-16087\\ P/Y=1\\ C/Y= \text{to solve}\\ $$

This gave me $k=1$, which results in the compounding interest to become $\$16068.38$ - which is incorrect. Any other methods to solve these kinds of questions - that is, where the x is in the denominator and the exponent - would be appreciated, thank you.

Answer 1

The result is quite close to $1$ but the equation cannot be solved even using special function.

If you make a series expansion $$0=-\frac{149}{8}+\frac{621}{4} \left(7+207 \log \left(\frac{207}{200}\right)\right)(x-1)+O\left((x-1)^2\right)$$ $$x\sim 1+\frac{149}{1242 \left(7+207 \log \left(\frac{207}{200}\right)\right)}=1.00850$$ while the "exact" solution is $1.00846$.

Using the approximate gives for $\text{(rhs-lhs)}$ a value of $0.087988$ instead of $-18.6250$ if $x=1$.

Answer 2

I'm not sure if you can solve this with a TVM calculator, but we can solve this iteratively. We have $$ f(x) = 15~000~\left(1 + \frac{3.5}{100x} \right)^{2x} - 16~087 $$ from which we are trying to find the root. We can use Newton's method, for example. The derivative is $$ f'(x)=30~000~\left(1+\frac{3.5}{100x}\right)^{2x} \cdot \frac{ (0.035 + x) \ln \left(\frac{0.035 + x}{x} \right) -0.035}{0.035 + x} $$

With a starting guess $x_0$, we can start iterating so that $$ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} $$ Let's start from $x_0=1$. Then we get the following table (rounding $x$ to integer values):

\begin{array}{|l|c|c|c|} \hline x_n & f(x_n) & f'(x_n) & x_{n+1} & \text{rounded} \\ \hline 1 & -18.625 & 18.800 & 1.991 & 2 \\ 2 & -9.115 & 4.811 & 3.894 & 4 \\ 4 & -4.275 & 1.217 & 7.512 & 8 \\ 8 & -1.833 & 0.306 & 13.990 & 14 \\ 14 & -0.783 & 0.100 & 21.809 & 22 \\ 22 & -0.272 & 0.041 & 28.697 & 29 \\ 29 & -0.056 & 0.023 & 31.406 & 31 \\ 31 & -0.013 & 0.020 & 31.611 & 32 \\ 32 & 0.007 & 0.019 & 31.619 & 32\\ \hline \end{array}

The iteration has converged to $x=32$.