Question

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.

Answer 1

Answer:

7.93

General Exponential Form: y=ab^x

Plug in both points

divide the equations

cancel out a, subtract exponent of b

Answer 2

Answer: $7.93$

Step-by-step explanation:

Given

$f(x)$ is an exponential function. Suppose $f(x)$ is $ae^{bx}$

$f(1.5)=5\ \text{and}\ f(7.5)=79$

$\Rightarrow 5=ae^{1.5b}\\\Rightarrow \ln 5=\ln a-1.5b\quad \ldots(i)$

Similarly,

$\Rightarrow 79=ae^{7.5b}\\\Rightarrow \ln(79)=\ln a+7.5b\quad \ldots(ii)$

Subtract (i) and (ii)

$\Rightarrow \ln (79)-\ln (5)=9b\\\Rightarrow \ln (\frac{79}{5})=9b\\\\\Rightarrow b=\dfrac{\ln (\frac{79}{5})}{9}\\\\\Rightarrow b=0.3066$

Insert the value of $b$

$\Rightarrow 5=ae^{0.46}\\\Rightarrow a=5\times 0.6312\\\Rightarrow a=3.156\approx 3.16$

So, the function becomes

$\Rightarrow f(x)=3.16e^{0.3066b}$

$\Rightarrow f(3)=3.16e^{0.3066\times 3}\\\Rightarrow f(3)=7.927\approx 7.93$