# Math Help - Logarithmic Functions

1. ## Logarithmic Functions

log2(x + 5) = 4 Find x.

log6(x2 - x) = 1 Find x.

log4x + log4(x + 6) = 2 Find x

log(x + 2) - log(x) = 2. Find x

Any help is appreciated. Thank You.

2. Originally Posted by twig684

log2(x + 5) = 4 Find x.
$2^4=x+5$

Originally Posted by twig684

log6(x2 - x) = 1 Find x.
$6^1=x^2-x$

Originally Posted by twig684

log4x + log4(x + 6) = 2 Find x
$\log_4(x(x+6))=2$

$4^2=x^2+6x$

Originally Posted by twig684

log(x + 2) - log(x) = 2. Find x
$\log\frac{x+2}{x}=2$

$10^2=\frac{x+2}{x}$

3. You didnt give any solutions...now Im discrediting your methods but I have the answers for each problem Im just trying to find out how to solve them..?

1. =11
2. =3 or -2
3. =2
4. =2/99

4. You didnt give any solutions...now Im discrediting your methods but I have the answers for each problem Im just trying to find out how to solve them..?
He gave you sufficient hints to help you do the problem. Did you try anything ?

Remember, in a test you won't have the answers, and you'll need the method to solve them ><

5. Wow for some reason I thought i got 9 with his first problem. Sorry my mistake.
But what I am asking for is just a whole run down of one problem so I can see exactly what he did.
Lets say in the 2nd problem I posted, how did he come up with the next step. I am not trying to be lazy but I cannot seem to find any examples in our book and my professor has yet to email me back.

6. Originally Posted by twig684
You didnt give any solutions...now Im discrediting your methods but I have the answers for each problem Im just trying to find out how to solve them..?
Merriam-Webster Online Dictionary

Main Entry: dis·cred·it
Pronunciation: \(ˌ)dis-ˈkre-dət\
Function: transitive verb 1 : to refuse to accept as true or accurate : disbelieve

My methods were just fine and accurate. All you have left to do is perform the algebra. Are you stuck on doing that?

7. 4^2=x^2+6x
So after taking the square root from each side, I still keep getting 4/7 which I know is not right. Yes I am stumped.

8. Originally Posted by twig684
4^2=x^2+6x
So after taking the square root from each side, I still keep getting 4/7 which I know is not right. Yes I am stumped.
$16=x^2+6x$

$x^2+6x-16=0$

$(x+8)(x-2)=0$

$x=-8 \ \ or \ \ x=2$

x cannot be negative, so x=2.

9. Thank you very much.