# How long an investment will take to compund to a target amount

A man with $\$20,000$to invest decides to diversify his investments by placing$\$10,000$ in an account that earns $7.2\%$ compounded continuously and $\$10,000$in an account that earns$8.4\%$compounded annually. Use graphical approximation methods to determine how long it will take for his total investment in the two accounts to grow to$\$35,000$.

Teacher said the answer is: $7.3$ years

But i don't know how to solve it.

FORMULAS:

Compounded Continuously: $A=Pe^{rt}$

Compounded Annually: $A=P(1+r)^t$

$A$ - Money after t years
$P$ - Principal amount invested
$r$ - rate of investment
$t$ - time in years
$e$ - the value $e$ used in $\ln$

• If it's graphical approximation methods, shouldn't your goal be to just graph it and estimate the numbers based off of that? It doesn't sound like your teacher is actually asking for a precision answer. – mardat Dec 4 '14 at 4:25
• You're looking for a value of $t$ that satisfies $$10000e^{0.072t}+10000(1.084)^t = 35000$$. – MPW Dec 4 '14 at 4:28
• Oh that's true @mardat but I don't know how to graph it to get the answer. – Audrey Dec 4 '14 at 4:34
• @MPW Can you factor out 10,000? When I did, I got the answer of 14.639, I think. How does one get 7.3? – Audrey Dec 4 '14 at 4:35
• @Audrey: Yes, then the equation becomes $$e^{0.072t} + 1.084^t =3.5$$ and it should have the same solution set. These equations can't be solved exactly, only by some approximation method. – MPW Dec 4 '14 at 4:43

The \$10,000 account at 7.2% interest gets you$10000e^{0.072t}$dollars in$t$years. The other \$10,000 investment gets you $10000(1+0.084)^t$ dollars in $t$ years.

So in $t$ years, you have $10000e^{0.072t}+10000(1+0.084)^t$ total dollars. We can make this a bit easier by factoring out the $10000$, so the amount of money in $t$ years, $M(t)$, we have is $M(t)=10000(e^{0.072t}+1.084^t)$.

We want to know in how many years will this be \$35,000. That is, we want$M(t)=35,000$. So $$10000(e^{0.072t}+1.084^t)=35,000$$ Divide by the 10,000 to get $$e^{0.072t}+1.084^t=3.5$$ Since your teacher wants this graphically, go into the calculator (I'll assume you have a TI-83 or TI-84) and graph the function$e^{0.072t}+1.084^t$as Y1. Then for Y2 3.5. Then when you graph them, look where they intersect. That$x$value where the lines touch will be the amount of years you want. Be careful about choosing an appropriate window! (But since we divided by the 10,000, this shouldn't be a problem - the only reason I did that). You can do this without having to 'guess' by using the 2nd - CALCULATE - then INTERSECT to find the intersection of the curves and make the calculator do all the work. To get an idea of what you should see, trying looking at this plot here. • Oh I see, this was really helpful. We don't use calculators in our classes though :( – Audrey Dec 4 '14 at 4:47 • If you aren't allowed calculators of some sort, you cannot solve this. Doing this graphically typically means using a graphing calculator. But if you only have a basic calculator, just graph this the way you would back when you first learned how to graph - plus in a value for$t$, get the value for$y$, plot the point, and repeat. Then connect the points, draw a horizontal line at$y=3.5$, then estimate your answer as best you can. If your calculator does not have$e$on it, remember$e\approx 2.718281828$. – mathematics2x2life Dec 4 '14 at 4:50 • @Audrey It is important that you can only solve this numerically, meaning you eventually have to use a calculator or computer of some sort. There is no exact numerical answer (or at least at the level of your class). – mathematics2x2life Dec 4 '14 at 4:52 • Alright. Thank you very much. I will try to get a hold of a calculator for this class. – Audrey Dec 4 '14 at 4:55 Everything you need is in the graphic per the instructions. As mentioned in the comments and shown in the graphic, you could also solve for$t$using numerical methods like Newton's Method. The equation becomes: $$10000e^{.072 t}+10000(1+0.084)^t = 35000 \implies e^{.072 t}+(1.084)^t = 3.5$$ The result is$t = 7.32505 ~~\mbox{years}$(see update below for more details). Here is the graphical representation (look at the point for the vertical line): Update We can solve this numerically (Fixed Point Iteration, Secant Method, Newton's Method ... Others). Using Newton's Method, we have the iteration formula: $$t_{n+1} = t_n- \dfrac{f(t)}{f'(t)} = t_n - \dfrac{e^{0.072 t_n} + 1.084^{t_n} -3.5}{0.0806579 \times 1.084^{t_n} + 0.072e^{0.072 t_n}}, ~ t_0 = -22$$ This converges in$13$steps to$t = 7.325046477477088$. • Alright, so how do you solve it numerically? We don't have the graphing calculators. – Audrey Dec 4 '14 at 4:47 • We weren't introduced to Newton's method. – Audrey Dec 4 '14 at 4:52 • Check with your teacher, but just judging by the nature of the question, and the fact that you don't know Newton's Method, I would assume it's something you don't need to know. – mardat Dec 4 '14 at 5:42 Using pencil, paper, eraser and a hand calculator that has exponential functions calculate and plot the graph, taking$1,000 as convenient unit :

$$A(n) = 10 ( 1 + .072)^n + 10 ( 1 + .084)^n$$

EDIT1:

corrected (oversight error) to include continuous compounding:

$$A(n) = 10\ e^{0.072 \, t} + 10\ ( 1 + .084)^n$$

for n upto 8 years. It takes ~7.32505 years to reach 35.