Use the base function f(x) = a * bx to solve for the bases a and b of the functions.

Question

Use the base function f(x) = a * b^x to solve for the bases a and b of the functions.

a. f(0) = 5, f(3) = 40                          b. f(0) = 3200, f(6) = 0.0032

c. f(3) = 12, f(5) = 48                        d. f(2) = 3,      f(4) = 15

Answer 1

Here are your answers:

(A) since f(0) = 5, your a in your formula would be 5 because a is just equal to the y-value when x = 0. Now for f(3) =40, is we plug this into f(x) = a(b^x) that would give us 40 = 5(b³). Dividing by 5 gives us 8 = x³. Now if we do (8)^1/3 = (b³)^1/3 , that would give us b = 2. Thus, your formula would be f(x) = 5(2^x)

The process is similar for part (b):

(B) f(0) = 3200, → a = 3200

f(6) = 0.0032 → 0.0032 = 3200(b⁶) → (divide both sides by 3200) → 0.001 = b⁶ → (0.000001)^1/6 = (b⁶)^1/6 → b= 0.1 → f(x) = 5(0.1)^x

Part (C) you have a little bit more to do:

f(3) = 12 → 12 = a(b³)

f(5) = 48 → 48 = a(b⁵)

Solve both sides for a and then set them equal to each other

12/b³ = a = 48/b⁵

Since they are both equal to a set them equal to each other:

12/b³ = 48/b⁵ → (multiply by b⁵) → 12b² = 48 → (divide both sides by 12) → b² = 4 → b =2

Thus, f(x) = a(2^x). Now plug one of the points into x and y, and then solve for a. I will choose the one with the smaller numbers, (3,12):

12 = a(2³) → 12 =a(8) → a = 12/8 → a = 3/2 or 1.5

Thus f(x) = 1.5(2^x)

Part (D) follows the same pattern as part (C):

Plug in the points to get 3 = a(b²) and 15 = a(b⁴).

Solve both for a to get a = 3/b² and a = 15/b⁴ .

Multiply both sides by b⁴ and then divide both sides by 3 to get b² = 5.

Take the square root of both sides to get b = √5

Now we have f(x) = a(√5^x)

Finally plug one of the points into x and y and then solve for a. I'm going to choose (2, 3), from f(2) = 3. This gives us:

3 = a(√5²) → 3 = a(5) → a = 3/5

Finally, our function is f(x) = (3/5)(√5^x)