An exponential function in the form y = ab^x goes through the points (3, 10.125) and (6, 34.2). Find a to the
nearest integer and b to the nearest tenth, then find f (10) to the nearest integer.

Respuesta :

Answer:

[tex]f(10) = 173[/tex]

Step-by-step explanation:

Given

Exponential Function

[tex](x_1,y_1) = (3,10.125)[/tex]

[tex](x_2,y_2) = (6,34.2)[/tex]

Required

Determine f(10)

We have that

[tex]y = ab^x[/tex]

First, we need to solve for the values of a and b

For [tex](x_1,y_1) = (3,10.125)[/tex]

[tex]10.125 = ab^3[/tex] --- (1)

For [tex](x_2,y_2) = (6,34.2)[/tex]

[tex]34.2 = ab^6[/tex] ---- (2)

Divide (2) by (1)

[tex]\frac{34.2}{10.125} = \frac{ab^6}{ab^3}[/tex]

[tex]\frac{34.2}{10.125} = \frac{b^6}{b^3}[/tex]

[tex]3.38= b^{6-3}[/tex]

[tex]3.38= b^{3}[/tex]

Take cube root of both sides

[tex]b = \sqrt[3]{3.38}[/tex]

[tex]b = 1.5[/tex]

Substitute 1.5 for b in [tex]10.125 = ab^3[/tex]

[tex]10.125 = a * 1.5^3[/tex]

[tex]10.125 = a * 3.375[/tex]

Solve for a

[tex]a = \frac{10.125}{3.375}[/tex]

[tex]a = 3[/tex]

To solve for f(10).

This implies that x = 10

So, we have:

[tex]y = ab^x[/tex] which becomes

[tex]y = 3 * 1.5^{10[/tex]

[tex]y = 3 * 57.6650390625[/tex]

[tex]y = 172.995117188[/tex]

[tex]y = 173[/tex] -- approximated

Hence:

[tex]f(10) = 173[/tex]