MHB Chris' question at Yahoo Answers regarding an exponential function

AI Thread Summary
To find the exponential function f(x) = ae^(bx) given f(0) = 3 and f(3) = 4, the value of a is determined to be 3. Substituting this into the second equation leads to the calculation of b, resulting in b = (1/3)ln(4/3). The final function can be expressed as f(x) = 3(4/3)^(x/3). This provides the required function in both exponential and simplified forms.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
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.
 
Mathematics news on Phys.org
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:

$$f(x)=3e^{\ln\left(\left(\frac{4}{3} \right)^{\frac{x}{3}} \right)}=3\left(\frac{4}{3} \right)^{\frac{x}{3}}$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top