If f(x)f(x) is an exponential function where f(1.5)=5f(1.5)=5 and f(7.5)=79f(7.5)=79, then find the value of f(3)f(3), to the nearest hundredth.

Respuesta :

Answer: [tex]7.93[/tex]

Step-by-step explanation:

Given

[tex]f(x)[/tex] is an exponential function. Suppose [tex]f(x)[/tex] is [tex]ae^{bx}[/tex]

[tex]f(1.5)=5\ \text{and}\ f(7.5)=79[/tex]

[tex]\Rightarrow 5=ae^{1.5b}\\\Rightarrow \ln 5=\ln a-1.5b\quad \ldots(i)[/tex]

Similarly,

[tex]\Rightarrow 79=ae^{7.5b}\\\Rightarrow \ln(79)=\ln a+7.5b\quad \ldots(ii)[/tex]

Subtract (i) and (ii)

[tex]\Rightarrow \ln (79)-\ln (5)=9b\\\Rightarrow \ln (\frac{79}{5})=9b\\\\\Rightarrow b=\dfrac{\ln (\frac{79}{5})}{9}\\\\\Rightarrow b=0.3066[/tex]

Insert the value of [tex]b[/tex]

[tex]\Rightarrow 5=ae^{0.46}\\\Rightarrow a=5\times 0.6312\\\Rightarrow a=3.156\approx 3.16[/tex]

So, the function becomes

[tex]\Rightarrow f(x)=3.16e^{0.3066b}[/tex]

[tex]\Rightarrow f(3)=3.16e^{0.3066\times 3}\\\Rightarrow f(3)=7.927\approx 7.93[/tex]

Answer:

7.93

General Exponential Form: y=ab^x

Plug in both points

divide the equations

cancel out a, subtract exponent of b