# Comparing Exponentials

Alignments to Content Standards: F-LE.A

1. Lincoln deposits money in a Certificate of Deposit account. The balance (in dollars) in his account $t$ years after making the deposit is given by $L(t)=500(1.05)^t$ for $t \geq 0$. Explain, in terms of the structure of the expression for $L(t)$, why Lincoln's balance can never be 499.
2. Helen deposits money in a similar Certificate of Deposit account. The balance in her account is described, with units as in Lincoln's deposit, $H(t)=600(1.04)^t$ for $t \geq 0$. Use the structure of the expressions for $L(t)$ and $H(t)$ to describe how the two balances compare over time.
3. By what percent does the value of $L(t)$ grow each year? What about $H(t)$? Explain.
4. During which year does Lincoln's balance to "catch up" with Helen's? Show your work.

## IM Commentary

This task gives students an opportunity to work with exponential functions in a real world context involving continuously compounded interest. They will study how the base of the exponential function impacts its growth rate and use logarithms to solve exponential equations.

## Solutions

Solution: 1

1. $(1.05)^0=1$ and positive powers of 1.05 are larger than 1, thus the minimum value $L(t)$ attains, if $t \geq 0$, is 500. From the point of view of the context, a CD account grows in value over time so with a deposit of \$500 the value will never drop to \$499.

2. The value of $H(t)$ at $t=0$ is greater than that of $L(t)$ at $t=0$. ($H(0)=600 > L(0)=500$). However the base of the expression for $L(t)$ is greater than the base of the expression for $H(t)$ ($1.05 > 1.04$). Multiplying 500 by 1.05 repeatedly will produce a greater value than that resulting from multiplying 600 repeatedly by 1.04. So, the values of $L(t)$ will eventually surpass the values of $H(t)$.

3. Recall that $L(t)=500\cdot 1.05^t.$ The CD value one year later is $L(t+1)$. Comparing these,
\begin{align} L(t+1)&=500(1.05)^{t+1}\\ &=500(1.05)^t\cdot(1.05)^1\\ &=L(t) \left(1+.05 \right)\\ &= L(t)(1) +L(t)(0.05)\end{align}
This indicates that over a period of one year, the value of $L(t)$ increases by 5%. The same method reveals that the value of $H(t)$ increases by 4% over one year. It is important to note here that exponential growth is characterized by increasing by equal factors over equal intervals. In this case, the intervals are periods of one year while the factors are $1.04$ for Helen's account and $1.05$ for Lincoln's account.

4. We set the expressions for $L(t)$ and $H(t)$ equal and solve for $t$: \begin{align} 500\cdot 1.05^t &= 600\cdot 1.04^t \\ \frac{500\cdot 1.05^t}{500\cdot 1.04^t} &= \frac{600\cdot 1.04^t}{500\cdot 1.04^t} \\ \frac{1.05^t}{1.04^t} &= \frac{600}{500} \\ \left(\frac{1.05}{1.04}\right)^t &= \frac{600}{500} \\ \ln \left(\frac{1.05}{1.04}\right)^t &= \ln \frac{600}{500} \\ t\ln \left(\frac{1.05}{1.04}\right) &= \ln \frac{600}{500} \\ t &= \frac{\ln \left(\frac{600}{500}\right)}{\ln \frac{1.05}{1.04}} \approx 19.0525 \text { years}\\ \end{align} So after $19$ years, Helen's balance is still larger than Lincoln's but after $20$ years, Lincoln's balance will be larger than Helen's. According to the calculation we just made, Lincoln's balance will be equal to Helen's balance very early in the $20^{\rm th}$ year.

Solution: 2 graphing technology for part (d)

Here we present a graph which shows Lincon's account balance and Helen's account balance over a period of $25$ years. Helen's account balance is larger for the first $19$ years and then shortly into the $20^{\rm th}$ year, the two accounts have equal balances, after which Lincoln's account will be larger:

So as in the first solution, we see that after 19 years, Helen's balance is still larger than Lincoln's but after 20 years, Lincoln's balance will be larger than Helen's.