# Math Help - Stuck on this log problem.

1. ## Stuck on this log problem.

f(x)=(Ln x)2- 3Ln x

x=2e
----------------------------------------------------
(ln(2e))2-3ln(2e)

2ln2e-3ln(2e)

im stuck :| logs and natural logs are the most confusing thing ever

2. ## Re: Stuck on this log problem.

$\ln(2e)^2 = 2\ln(2e)$ is not correct.

What is correct is

$\ln((2e)^2)=2\ln(2e)$

$f(x)=(\ln(x))^2-3\ln(x)$

$f(2e)=(\ln(2e))^2 - 3\ln(2e)$

this is actually in fairly simple form. You could further do the following

$u=\ln(2e)$

$f(2e)=u^2-3u = (u-1.5)^2-2.25 = (\ln(2e)-1.5)^2-2.25$

3. ## Re: Stuck on this log problem.

Originally Posted by romsek
$\ln(2e)^2 = 2\ln(2e)$ is not correct.

What is correct is

$\ln((2e)^2)=2\ln(2e)$

$f(x)=(\ln(x))^2-3\ln(x)$

$f(2e)=(\ln(2e))^2 - 3\ln(2e)$

this is actually in fairly simple form. You could further do the following

$u=\ln(2e)$

$f(2e)=u^2-3u = (u-1.5)^2-2.25 = (\ln(2e)-1.5)^2-2.25$
I think you read my answer wrong hahaa, could you also go step by step?

but i was given these answer choices:
1. f(2e) = 3(ln 2)2+ ln 2 + 2
2. f(2e) = 3(ln 2)2+5 ln 2 + 4
3. f(2e) = (ln 2)2− ln 2 − 2
4. f(2e) = (ln 2)2+ ln 2 + 2
5. f(2e) = 3(ln 2)2− ln 2 − 2
6. f(2e) = (ln 2)2−5 ln 2 − 4
7. f(2e) = 3(ln 2)2−5 ln 2 − 4
8. f(2e) = (ln 2)2+5 ln 2 + 4

4. ## Re: Stuck on this log problem.

Originally Posted by chewydrop
I think you read my answer wrong hahaa, could you also go step by step?

but i was given these answer choices:
1. f(2e) = 3(ln 2)2+ ln 2 + 2
2. f(2e) = 3(ln 2)2+5 ln 2 + 4
3. f(2e) = (ln 2)2− ln 2 − 2
4. f(2e) = (ln 2)2+ ln 2 + 2
5. f(2e) = 3(ln 2)2− ln 2 − 2
6. f(2e) = (ln 2)2−5 ln 2 − 4
7. f(2e) = 3(ln 2)2−5 ln 2 − 4
8. f(2e) = (ln 2)2+5 ln 2 + 4
is $e$ some variable or is it the $e$ that's the base of the natural log?

5. ## Re: Stuck on this log problem.

Originally Posted by romsek
is $e$ some variable or is it the $e$ that's the base of the natural log?
i believe x=2e is just some variable.

6. ## Re: Stuck on this log problem.

Originally Posted by chewydrop
i believe x=2e is just some variable.

7. ## Re: Stuck on this log problem.

$f(2e)=(\ln(2e))^2 - 3\ln(2e)$

$\ln(2e)=\ln(2)+\ln(e)=1+\ln(2)$

so we have

$f(2e)=(1+\ln(2))^2-3(1+\ln(2))=$

$1+2\ln(2)+(\ln(2))^2-3-3\ln(2)=$

$-2 - \ln(2) + (\ln(2))^2$

which matches choice (3)