Chris' question at Yahoo Answers regarding an exponential function

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The discussion focuses on deriving an exponential function of the form f(x) = ae^(bx) given specific function values. The values provided are f(0) = 3 and f(3) = 4. By substituting f(0) into the equation, it is determined that a = 3. Using this value, the second point leads to the calculation of b as b = (1/3)ln(4/3). The final function is expressed as f(x) = 3(4/3)^(x/3), which simplifies the original exponential form.

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MarkFL
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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.
 
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Hello chris,

We are given:

$$f(x)=ae^{bx}$$

where:

$$f(0)=3,=,f(3)=4$$

Using the first point, we find:

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

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

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

And so our function is:

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

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

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

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