# Exponential Functions

• Jan 11th 2009, 08:07 AM
Macleef
Exponential Functions
Quote:

Once the initial publicity surrounding the release of a new book is over, sales of the hardcover edition tend to decrease exponentially. At the time publicity was discontinued, a certain book was experiencing sales of 25,000 copies per month. One month later, sales of the book had dropped to 10,000 copies per month. What will the sales be after 1 more month?
I have no idea on how to solve this problem. Please show me a step by step solution! The answer is 4000.

And also need assistance with the following question...

Find $f(2)$ if $f(x) = e^{kx}$ and $f(1) = 20$.

Here's what I did...

$20 = e^{k \times 1}$
$e^{20} = e^{k \times 1}$
$20 = e$

$f(2) = e^{20 \times 2}$
$f(2) = 2.3539$ x $10^{17}$

• Jan 11th 2009, 08:13 AM
james_bond
For your second question: $20=f(1)=e^k\rightarrow k=\log 20$ so $f(2)=e^{2\cdot \log 20}=\left(e^{\log 20}\right)^2=20^2=\boxed{400}$.
• Jan 11th 2009, 08:13 AM
Prove It
Quote:

Originally Posted by Macleef
I have no idea on how to solve this problem. Please show me a step by step solution! The answer is 4000.

And also need assistance with the following question...

Find $f(2)$ if $f(x) = e^{kx}$ and $f(1) = 20$.

Here's what I did...

$20 = e^{k \times 1}$
$e^{20} = e^{k \times 1}$
$20 = e$

$f(2) = e^{20 \times 2}$
$f(2) = 2.3539$ x $10^{17}$

Step 1 to Step 2 is wrong. How have you turned 20 into $e^{20}$?

You should have

$20 = e^k$

$k = \ln{20}$.

Therefore $f(x) = e^{x\ln{20}} = e^{\ln{20^x}} = 20^x$.

Thus $f(2) = 20^2 = 400$.
• Jan 11th 2009, 08:19 AM
james_bond
Quote:

Originally Posted by Macleef
$20 = e^{k \times 1}$
$e^{20} = e^{k \times 1}$
$20 = e$

$f(2) = e^{20 \times 2}$
$f(2) = 2.3539$ x $10^{17}$

How did you get that?? If you take a look at the first two lines you can see that it just can be true. (It would mean that $20=e^{20}$..) I suppose that you meant $20 = k$ in the third line, then it would at least come from the line above and the other two lines would come from that...