# Growth rate

• Oct 29th 2012, 06:45 PM
Eraser147
Growth rate
Suppose that the number of cell phones in the world increases by a total of 160% over a five-year period. What is the continuous growth rate for the number of cell phones in the world?
• Oct 29th 2012, 08:33 PM
chiro
Re: Growth rate
Hey Eraser147.

Continuous growth is exponential where you are looking at A*e(bx) for constant A and b. You need two values to solve for two independent equations but you know f(x+5) = 1.6*f(x) for some particular value of x and you can cancel A from both sides.

Can you use this to solve for b?
• Oct 29th 2012, 08:42 PM
Eraser147
Re: Growth rate
1.6(x+5) = 1.6x+8? b=8??
• Oct 29th 2012, 08:48 PM
chiro
Re: Growth rate
No, f(x) is a function: f(x) = Ae^(bx) where f(x+5) = Ae^(b[x+5]).

You are not multiplying a constant f by x: f is a function that takes a value for x and spits out a value called f(x).

1.6*f(x+5) = 1.6*A*e(b[x+5]) = 1.6Ae^[bx + 5b] = 1.6Ae^[bx]*e^[5b] = f(x).
• Oct 29th 2012, 08:55 PM
Eraser147
Re: Growth rate
Sorry, but I really don't know where to go from 1.6Ae^[bx]*e^[5b] = f(x). Can you please show me the whole entire solution. I think that would help me trace out everything that you are trying to tell me.
• Oct 29th 2012, 09:00 PM
chiro
Re: Growth rate
What does f(x) equal? (Hint: I posted it above, just plug it in).
• Oct 29th 2012, 09:07 PM
Eraser147
Re: Growth rate
Well, I just figured it out this way which was similar to my student's solution manual.. If the initial amount was 0, then... if it raised up to 160%, that means that it went up 2.6 times the original number. Then I can have this formula 2.6P=Pe^5r which is ln2.6/5
• Oct 29th 2012, 09:09 PM
chiro
Re: Growth rate
Yes sorry I made a mistake in that you should have 2.6f(x+5) = f(x). If you cancel out the A and e^(bx) from both sides you will get exactly the same equation as in the solution manual.