Answer:
3a. $8493.42
3b. 139.0 months
4a. 10
4b. 10.2 years
Explanation:
Given exponential equations, you want values for specific times, and you want times for specific values.
In part (a), the value is found by substituting an appropriate value for t and doing the arithmetic.
In part (b), logarithms will be involved in solving for t.
3. Investment
(a) Use t=12, the number of months in one year. You want the value of ...
S = 8000·1.005^12 ≈ 8493.42
The amount after 1 year is $8493.42.
(b) The time it takes to double the investment is the value of t such that ...
16000 = 8000·1.005^t
2 = 1.005^t . . . . . . divide by 8000
log(2) = t·log(1.005) . . . . . take logarithms
t = log(2)/log(1.005) ≈ 138.976
The investment will double in about 139.0 months.
4. Otters
(a) Use t=0 and evaluate.
y = 2500 -2490e^(-0.1·0) = 2500 -2490 = 10
The population of otters was 10 when they were reintroduced.
(b) The time it takes for the population to reach 1600 is the value of t such that ...
1600 = 2500 -2490e^(-0.1t) . . . . use 1600 for y
-900 = -2490e^(-0.1t) . . . . . . . . . subtract 2500
900/2490 = e^(-0.1t) . . . . . . . . . divide by -2490
ln(900/2490) = -0.1t . . . . . . . . . take natural logs
t = ln(900/2490)/-0.1 . . . . . . . . divide by the coefficient of t
t ≈ 10.176 ≈ 10.2
It will be 10.2 years before the otter population reaches 1600.