Solutions
Solution:
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.

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 $H(t)$.

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.

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.