1. (a) e = 4e

```1. (a)
e6x = 4e12
e6x
=4
e12
e6x−12 = 4
ln e6x−12 = ln 4
6x − 12 = ln 4
6x = 12 + ln 4
ln 4
x=2+
6
x ≈ 2.23105
(b)
log2 (x − 1) + log2 (x − 2) = log2 3 − 2
log2 (x − 1) + log2 (x − 2) − log2 3 = −2
(x − 1)(x − 2)
log2
= −2
3
(x−1)(x−2)
) = 2−2
3
2log2 (
(x − 1)(x − 2)
1
=
3
4
3
(x − 1)(x − 2) =
4
3
x2 − 3x + 2 =
4
5
x2 − 3x + = 0
4
q
3&plusmn; 9−4&middot;1&middot;
x=
5
4
2
1
5
x = or x =
2
2
Checking these two answers, we find that x = 12 does not work– we end up taking
logs of negative numbers. So our only answer is x = 52 .
2. (a) At t = 0, there are 12 rabbits. At t = 3, the number has doubled, so there are 24
rabbits.
(b) We will use the points (0,12) and (3,24) to solve for the constants a and k.
12 = aek&middot;0
12 = a &middot; 1
a = 12
Using the other point and this value for a:
24 = 12ek&middot;3
2 = e3k
ln 2 = ln e3k
ln 2 = 3k
ln 2
k=
3
k ≈ .23105
(c)
2000 = 12e.23105t
2000
= e.23105t
12
2000
ln
= ln e.23105t
12
2000
= .23105t
ln
12
ln 2000
12
t=
.23105
t ≈ 22.23
So after 22.23 days, there will be 2000 rabbits.
(d)
y = 12e.23105&middot;40
= 123, 859.935
So there will be 123,860 rabbits.
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3. (a) Present value of an annuity due
(b) APY of an account
(c) Sinking fund
(d) Periodic payment of an amortized loan
(e) Simple interest
(f) Future value of an ordinary annuity
(g) Continuously compounded interest
(h) Loan payoff amount
(i) Present value of an ordinary annuity
(j) Total interest paid
(k) Future value of an annuity due
(l) Periodically compounded interest
(m) Present value of a deferred annuity
Page 3
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