SOLUTION: Two die- cast metal car models were purchased at the same time. The first cost $99.50 and appreciated by 10% yearly. The second cost $49.50 and appreciated 16% yearly. Algebraicall

Question 127471: Two die- cast metal car models were purchased at the same time. The first cost $99.50 and appreciated by 10% yearly. The second cost $49.50 and appreciated 16% yearly. Algebraically determine how many years will it take for the two models to be of equal value.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
The first cost $99.50 and appreciated by 10% yearly. The second cost $49.50 and appreciated 16% yearly. Algebraically determine how many years will it take for the two models to be of equal value.
:
Use the annual interest equation, t = time in years
:
Cheap car value = expensive car value
:
49.5(1.16)^t = 99.5(1.10)^t
:
Using natural logs:
ln(49.5) + ln(1.16^t) = ln(99.50) + ln(1.10^t)
:
log equiv of exponents:
ln(49.5) + t*ln(1.16) = ln(99.50) + t*ln(1.10)
:
3.902 + .1484t = 4.6 + .0953t
:
.1484t - .0953t = 4.6 - 3.902
:
.0531t = .698
t =
t = 13.14 years
;
:
Find the value of each after 13.14 yrs, using a calc
1.16^13.14 * 49.5 = $348
1.1^13.14 * 99.5 = $348 confirms our solution
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