Answer to Question #147914 in Algebra for Bonevie tutor

Question

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?

Answer 1

1. A = P (1 + r/n) ^(nt)

16500 = 5500(1+0.08/2)^2t

16500 = 5500(1.04)^2t

16500/5500 = 1.04^2t

3= 1.04^2t

Take log of both sides

Log3 = 2t log 1.04

2t = log3/log 1.04

2t = 28

t = 14 years

2. A(t) = ab^t 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