Find C and a so that f(x)=Ca^x satisfies the given equation.

Question

Find C and a so that f(x)=Ca^x satisfies the given equation.

 

f(-2) = 25 ,      f(2) = 1

           3                  75

 

a = ?

Answer 1

Hi Keith;

Let's begin with...

f(2)=ca²=1/75=(1/3)(1/25)=(1/3)(1/5)²

c=1/3

a=1/5

 

f(-2)=ca^-2=c/a²=25/3

f(-2)=(1/3)(1/5)^-2=(1/3)(5)²=25/3

 

 

Answer 2

f(x) = C a^x where a > 0 by definition of exponential function.

f(2) = 1/75 = C a^2

f(-2) = 25/3 = C a^(-2)

f(2) * f(-2) = (1/75) (25/3) = (C a^2) (C a^(-2))

(25/(3(25))) (1/3) = C^2 a^(2 + -2)

(1/3) (1/3) = C^2 a^(0)

1/3 = C

1/75 = (1/3) a^2

3/(3*25) = a^2

(1/5)^2 = a^2

a = 1/5

f(x) = (1/3) (1/5)^x

check:

f(2) = (1/3) (1/5)^2 =? 1/75

(1/3) (1/25) = 1/75 check!

f(-2) = (1/3) (1/5)^(-2) =? 25/3

(1/3) (5)^(2) =? 25/3

25/3 = 25/3 check!

Answer 3

This is a problem of two equations and two unknowns.

 

First solve the equation for the two given values:

 

f(-2) = 25/3 = Ca^-2 and f(2) = 1/75 = Ca²

25/3 = Ca^-2

25/3 = C/a²

cross multiply

25a² = 3C

a² = 3C/25

a = ± square root(3C/25)

 

and

 

f(2) = 1/75 = Ca²

1/75*C = a²

1/5*square root(3C) = a

 

thus a=a =>

square root(3C/25) = 1/5*square root(3C)

square root(3C)/5 = 1/5*square root(3C)

cross multiply:

(square root(3C))^{2 = 5/5}

3C = 1

C = 1/3

 

Use either equations given:

f(-2) = 25/3 = (1/3)a^-2

25/3 = 1/3a²

cross multiply:

75a² = 3

25a² = 1

a² = 1/25

a = 1/5

 

QED