# Natural Logarithms Help Plz!

• Feb 26th 2008, 12:39 PM
topaz192
Natural Logarithms Help Plz!
I managed to do all of my homework except this one:

How do you solve the equation: ln x - ln 5= -1??
• Feb 26th 2008, 12:43 PM
Peritus
$\begin{gathered}
\ln x - \ln 5 = - 1 \hfill \\
\ln \frac{x}
{5} = - 1 \hfill \\
\Leftrightarrow \frac{x}
{5} = e^{ - 1} \hfill \\
\Leftrightarrow x = 5e^{ - 1} \hfill \\
\end{gathered}
$
• Feb 26th 2008, 12:46 PM
topaz192
how did you do that? You divide the numbers x and 5 together?

And after that I just find out what e^-1 is?
• Feb 26th 2008, 12:50 PM
Jhevon
Quote:

Originally Posted by topaz192
how did you do that? You divide the numbers x and 5 together?

it is a rule

$\log_a x - \log_a y = \log_a \frac xy$

Quote:

And after that I just find out what e^-1 is?
no. leave it as e^{-1}. or you can write the answer as 5/e if you wish
• Feb 26th 2008, 12:54 PM
topaz192
ok i thought e cant get left in the answer?
• Feb 26th 2008, 01:05 PM
Jhevon
Quote:

Originally Posted by topaz192
ok i thought e cant get left in the answer?

it can...unless the problem specifically asked you for your answer in decimal
• Feb 26th 2008, 01:09 PM
topaz192
yea it did. so how do you get an answer? isnt e^-1?
• Feb 26th 2008, 01:10 PM
Jhevon
Quote:

Originally Posted by topaz192
yea it did. so how do you get an answer? isnt e^-1?

um, just plug it into your calculator
• Feb 26th 2008, 01:13 PM
topaz192
ok! thanks for your time! = )
• Feb 26th 2008, 01:14 PM
Jhevon
Quote:

Originally Posted by topaz192
ok! thanks for your time! = )

you're welcome :)