Chris' question at Yahoo Answers regarding an exponential function

[MarkFL]

Here is the question:

Find a function of the form f(x)=aebx given the function values? Please help!?

f(0)=3, f(3)=4 ; how can I solve this? I keep on getting the wrong answers..

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello chris,

We are given:

\(\displaystyle f(x)=ae^{bx}\)

where:

\(\displaystyle f(0)=3,=,f(3)=4\)

Using the first point, we find:

\(\displaystyle f(0)=ae^{b\cdot0}=a=3\)

Now, using this value for $a$, and the second point, we find:

\(\displaystyle f(3)=3e^{b\cdot3}=4\implies e^{3b}=\frac{4}{3}\implies b=\frac{1}{3}\ln\left(\frac{4}{3} \right)\)

And so our function is:

\(\displaystyle f(x)=3e^{\frac{1}{3}\ln\left(\frac{4}{3} \right)x}\)

Although this is the form required, we could rewrite it in a simpler form:

\(\displaystyle f(x)=3e^{\ln\left(\left(\frac{4}{3} \right)^{\frac{x}{3}} \right)}=3\left(\frac{4}{3} \right)^{\frac{x}{3}}\)