# Solving $5000 = 2000(1 + (0.0225/12))^{12t} + 2000(1 + (0.03/12))^{12t}$

So I recently had a math problem in my algebra 2 class:

There are two accounts with a principle of $2000$, compounded monthly, one growing at $3\%$ and one growing at $2.25\%$. Find when the sum of both accounts is $5000$.

I came up with a relatively simple equation: $$5000 = 2000(1 + (0.0225/12))^{12t} + 2000(1 + (0.03/12))^{12t}.$$

When I went to use Logarithms to solve this, I became frustrated upon not finding a single way to isolate both $t'$s. I asked several of my friends in higher math classes, and none of them had a clue. I did some research and after simplifying my equation to $$\log_{(1 + (0.0225/12))}{\left(\frac{5} {2} - (1 + (0.03/12))^{12t}\right)} = 12t$$

I was able to find the identity $$\log{(a + b)} = \log{a} + \log\left({1 + \frac{b}{a}}\right).$$

Which would be able to transform the left equation into something which again has two terms, although one of those terms would be a $1$. I'm not even sure this would be particularly helpful, but it was all I could find in my research. I know the answer to $t$ is $\approx 8.49$, although this was only done with a graphing calculator. I tried Wolfram Alpha, but it didn't even show the answer to the equation.

So far I've talked to two teachers, my algebra 2 teacher, and the BC calculus teacher (who I talk to a lot), and both of them told me that it is only possible through graphing and seeing where the constant equation and the equation of the addition of the exponential formulas intersect. To me this doesn't make any sense though, the solution is available and doesn't look like it would require any iteration like some other problems. Is it true that it is simply impossible to solve this algebraically, or is there some way to solve this?

• Problems like this seldom have simple, analytic solutions. Numerical methods are, by far, the best. If you want an approximate method, note that this problem is roughly the same as saying you invest $\$4000$at the average rate of$2.625$. That one you can solve analytically and you get$8.51$...not bad for a crude approximation. – lulu Feb 26, 2018 at 22:35 • That's frustrating... are there any proofs that numerical methods are necessary, or are they simply used because finding a method of simplifying the answer is unlikely? Feb 26, 2018 at 22:46 • It's so unlikely that people who work in financial mathematics just default to numerical methods without a second thought. To be sure, there are a few cases that admit good analytic theories and these are studied intensely (if nothing else, perturbing those models gives you a good way to start your numerical analysis). – lulu Feb 26, 2018 at 22:50 ## 3 Answers The first point is to name $$\bigg \{ \begin{matrix} C_1= 1+\frac{0,0225}{12} \\ C_2= 1+\frac{0,03}{12}\end{matrix}$$ and your equation becomes $$C_1^{\ \ \ 12t} + C_2^{\ \ \ 12t} = \frac{5}{2}$$ Now, let's call $$m_1=log(C_1)$$ and $$m_2=log(C_2)$$ and the equation can be written as $$x^{m_1} + x^{m_2} = 2.5$$ where $$x=e^{12\cdot t}$$. Until now, it has nothing new. In the last two years, however, I've been working in a function proposed by R. V. Ramos named Lambert-Tsallis function, $$W_q(z)$$, https://doi.org/10.1016/j.physa.2019.03.046. In the original article , he proposed $$W_q(z)$$ as solution of the problem $$W_q(z) \cdot \bigg( 1 + (1-q)\cdot W_q(z) \bigg)^\frac{1}{1-q} = z$$ For sake of simplicity, I will change the index $$q$$ by $$r$$, and refer to the function as $$W_r(z)$$ which is the solution of $$W_r(z) \cdot \bigg( 1 + \frac{W_r(z)}{r} \bigg)^r = z$$ where $$r=1/(q-1)$$. From now on I will refer to subindex "r" always. After a few manipulations you will find $$x=\bigg [ \frac{1}{r} \cdot W_r(r \cdot 2.5^r) \bigg]^{\frac{1}{m_2 - m_1}}$$ where $$r=\frac{m_2 -m_1}{m_1}$$ Substituting your values will produce $$r \approx 1/3$$ and $$t=\frac{log(x)}{12}$$ This approach produces the result $$t=8.499994853871394$$, analytically. As already said in comments, you will need some numerical method but you can easily find the range in which the solution is since if $$f= 2000(1 + (0.0225/12))^{12t} + 2000(1 + (0.0300/12))^{12t}$$ $$g= 2000(1 + (0.0225/12))^{12t} + 2000(1 + (0.0225/12))^{12t}=4000(1 + (0.0225/12))^{12t}$$ $$h= 2000(1 + (0.0300/12))^{12t} + 2000(1 + (0.0300/12))^{12t}=4000(1 + (0.0300/12))^{12t}$$ you have $$g < f < h$$ Solving$5000=g$and$5000=h$does not make any problem and leads to $$7.44741 < t <9.92679$$ So, select for$t_0\$ some value between these bounds (the average for example, and start Newton method which would converge quite fast as shown below $$\left( \begin{array}{cc} n & t_n \\ 0 & 8.687099035 \\ 1 & 8.491460917 \\ 2 & 8.490944850 \\ 3 & 8.490944847 \end{array} \right)$$ In practice, you do not need such accuracy and, more than likely, one or two iterations would be sufficient.

With the abbreviations from the answer of K Z Nobrega, we get the following.

$$x^{m_1}+x^{m_2}=\frac{5}{2}$$

This equation is related to an equation similar to a trinomial equation.

$$x^{m_1}+x^{m_2}-\frac{5}{2}=0$$ $$\frac{2}{5}x^{m_1}+\frac{2}{5}x^{m_2}-1=0$$ $$x\to\left(\frac{5}{2}z\right)^\frac{1}{m1}$$: $$z-\left(-\left(\frac{5}{2}\right)^{\frac{m2}{m1}-1}\right) z^\frac{m2}{m1}-1=0$$

Now the equation is in the form of equation 8.1 of [Belkic 2019]. Solutions in terms of Bell polynomials, Pochhammer symbols or confluent Fox-Wright Function $$\ _1\Psi_1$$ can be obtained therefore.
$$\$$

Szabó, P. G.: On the roots of the trinomial equation. Centr. Eur. J. Operat. Res. 18 (2010) (1) 97-104

Belkić, D.: All the trinomial roots, their powers and logarithms from the Lambert series, Bell polynomials and Fox–Wright function: illustration for genome multiplicity in survival of irradiated cells. J. Math. Chem. 57 (2019) 59-106