# exponential

• Jul 12th 2008, 10:19 AM
norwoodjay
exponential
Dissolving a chemical; The amount of a chemical that will dissolve in asolution increases exponentially as the (Celsius) temperature t is increased according to the model:
A(t)=10e .0095t

At what temperature will 5 g dissolve?
• Jul 12th 2008, 10:36 AM
CaptainBlack
Quote:

Originally Posted by norwoodjay
Dissolving a chemical; The amount of a chemical that will dissolve in asolution increases exponentially as the (Celsius) temperature t is increased according to the model:
A(t)=10e .0095t

At what temperature will 5 g dissolve?

This requires that you solve the equaltion:

$5=10e^{0.0095t}$

or:

$1=2e^{0.0095t}$

you do this by first taking natural logs to get:

$\ln(1)=\ln(2)+0.0095t$

Now solve this linear equation

RonL
• Jul 12th 2008, 09:58 PM
norwoodjay
I need the final answer I am kind of confused.
• Jul 13th 2008, 05:25 AM
topsquark
Quote:

Originally Posted by norwoodjay
I need the final answer I am kind of confused.

What, specifically, do you need help on? CaptainBlack's derivation or solving the equation from the point where he left off?

-Dan
• Jul 13th 2008, 08:32 AM
norwoodjay
Yes were you left off from.
• Jul 13th 2008, 11:07 AM
janvdl
Quote:

Originally Posted by norwoodjay
Yes were you left off from.

Quote:

Originally Posted by CaptainBlack
This requires that you solve the equaltion:

$5=10e^{0.0095t}$

or:

$1=2e^{0.0095t}$

you do this by first taking natural logs to get:

$\ln(1)=\ln(2)+0.0095t$

Now solve this linear equation

RonL

$\ln(1) = 0$

$\ln(2) = 0.693147$

So we have:

$0.0095t = -0.693147$

$t = \frac{-0.693147}{0.0095}$