1. ## logarithms help.

okay, here's my problem.

ln 4 + ln x = 3

i am enrolled in an independent study course and i am having a lot of trouble with this problem. i am completely unsure as of how to go about solving this problem. there are other problems i need help with as well. but maybe help with this one will help me figure out the others.

any help offered would be appreciated!
thank you. =]

2. Originally Posted by beli3ve
okay, here's my problem.

ln 4 + ln x = 3

i am enrolled in an independent study course and i am having a lot of trouble with this problem. i am completely unsure as of how to go about solving this problem. there are other problems i need help with as well. but maybe help with this one will help me figure out the others.

any help offered would be appreciated!
thank you. =]
$\displaystyle \ln 4 + \ln x = 3$

$\displaystyle \Rightarrow \ln 4x = 3$ ..........since $\displaystyle \log_a xy = \log_a x + \log_a y$

$\displaystyle \Rightarrow 4x = e^3$ ..........since if $\displaystyle \log_a b = c$ then $\displaystyle a^c = b$. Note that $\displaystyle \ln \equiv log_e$

$\displaystyle \Rightarrow x = \frac {e^3}{4}$

3. Originally Posted by Jhevon
$\displaystyle \ln 4 + \ln x = 3$

$\displaystyle \Rightarrow \ln 4x = 3$ ..........since $\displaystyle \log_a xy = \log_a x + \log_a y$

$\displaystyle \Rightarrow 4x = e^3$ ..........since if $\displaystyle \log_a b = c$ then $\displaystyle a^c = b$. Note that $\displaystyle \ln \equiv log_e$

$\displaystyle \Rightarrow x = \frac {e^3}{4}$
I am having a hard time understand
$\displaystyle \Rightarrow x = \frac {e^3}{4}$.

4. Wait, I am getting now what ln is equivalent to, rather than ln = log, I believe that's where my mistake was. But now at the same time, it was only a 3.

How/why did it change to $\displaystyle e^3$?

5. Originally Posted by beli3ve
I am having a hard time understand
$\displaystyle \Rightarrow x = \frac {e^3}{4}$.
i had 4x = e^3 so i just divided both sides by 4, that's all

6. Originally Posted by beli3ve
Wait, I am getting now what ln is equivalent to, rather than ln = log, I believe that's where my mistake was. But now at the same time, it was only a 3.

How/why did it change to $\displaystyle e^3$?
as i told you, there is a law of logarithms that says:

if $\displaystyle \log_a b = c$ then $\displaystyle a^c = b$

we had $\displaystyle \ln 4x = 3$ we want to find x, so i applied the above law. since we can think of ln as log to the base e, my next line holds true

so it is like we had $\displaystyle \log_e (4x) = 3$ so $\displaystyle e^3 = 4x$

7. I wasn't making the connection of that last law.
Thank you very much. =]

That has made my next couple problems much easier.
You're a blessing.
Have a great rest of the day. =]