Solving word problem involving logarithmic functions, equations and inequalities
Kevin invested 5,500 pesos at an interest rate of 8% compounded semi-annually, how long will it take for the investment to reach 16,500 pesos?
The population of a small town growns exponentially at a rate of 5% that was 12,000 in 1993 and 21,000 in 1998. In what year will the population reach 36,750?
1. A = P (1 + r/n) (nt)
16500 = 5500(1+0.08/2)2t
16500 = 5500(1.04)2t
16500/5500 = 1.042t
3= 1.042t
Take log of both sides
Log3 = 2t log 1.04
2t = log3/log 1.04
2t = 28
t = 14 years
2. A(t) = abt where t is number of years after 1993.
Solve for a and b using (0,12000) and (5,21000).
Using 0,12000 you get: 12000 = ab^0
So a = 12000
Equation:
A(t) = 12000*b^t
Using (5,21000), solve for "b":
21000 = 12000*b^5
b^5 = 21/12 = 7/4
b = (7/4)^(1/5)
Equation:
A(t) = 12000*(7/4)^(t/5)
In what year will the population reach 36 750?
Solve:
36750 = 12000(7/4)^(t/5)
(7/4)^(t/5) = 3.0625
Take the log to solve for "t":
(t/5)log(7/4) = log(3.0625)
(t/5) = log(3.0625)/log(7/4)
t = 1.68 years
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