# Trying to figure this out?

• Dec 4th 2010, 11:07 AM
Trying to figure this out?
We were given 1 example to do math questions. But I don't get how I got the answer.

With with Log and Ln, but in this one it's Ln.
2.6=1.4e^(0.3t)
I have to solve for t.
So you have to divide both sides by 1.4 and add Ln as well.
Ln(2.6/1.4) = Ln e^(0.3t)
Now my example in my book does not make sense. I did not make a note next to it.
What are the next examples to get to the answer of t=2.06.

If someone would kindly help me out that would be appreciated.
Thanks
Joanne
• Dec 4th 2010, 11:19 AM
tonio
Quote:

We were given 1 example to do math questions. But I don't get how I got the answer.

With with Log and Ln, but in this one it's Ln.
2.6=1.4e^(0.3t)
I have to solve for t.
So you have to divide both sides by 1.4 and add Ln as well.
Ln(2.6/1.4) = Ln e^(0.3t)
Now my example in my book does not make sense. I did not make a note next to it.
What are the next examples to get to the answer of t=2.06.

If someone would kindly help me out that would be appreciated.
Thanks
Joanne

You don't "add" log or ln to both sides: you apply it to both sides, and that only if you're sure they are positive, so:

$\displaystyle{\ln\left(\frac{2.6}{1.4}\right)=\ln( e^{0.3t})=0.3t\Longrightarrow t=\frac{\ln\left(\frac{2.6}{1.4}\right)}{0.3}}$ , as

$\ln e^x=x$ since $e^x\,,\ln x$ are inverse function to each other.

Tonio
• Dec 4th 2010, 11:20 AM
emakarov
The functions $\ln x$ and $e^x$ are inverse, i.e., $\ln(e^x)=x$ for all x and $e^{\ln x}=x$ for all $x>0$. Therefore, $\ln(2.6/1.4) = \ln e^{0.3t}=0.3t$, from which t can be found.
• Dec 4th 2010, 11:20 AM
pickslides
You are on the right track,

$2.6 = 1.4 e^{0.3t}$

$\frac{2.6}{1.4}= e^{0.3t}$

$\ln \frac{2.6}{1.4}= 0.3t$

$\displaystyle\frac{\ln \frac{2.6}{1.4}}{0.3}= t$
• Dec 4th 2010, 11:29 AM
OK I get that now. But what happens to the e on the 2nd line?? Is that the e that is applied to the left side as Ln??? Do I have that correct?

Thank you all for the help to clear this up.
Joanne

OK I called a friend of mine and the Ln is e with a base of e, so the answer is 1. So it's not needed in a way, correct?
• Dec 4th 2010, 02:38 PM
pickslides
They are the inverse of each other.

I.e $x+3-3 = x-3+3 = x$

$\displaystye \ln e^x = e^{\ln x} = x$

and also,

$\displaystye \log_{a} a^x = a^{\log_{a} x} = x$