# Logarithmic Functions

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• Nov 5th 2008, 01:05 PM
twig684
Logarithmic Functions
Cant figure out these 4: Please help.

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.
• Nov 5th 2008, 01:25 PM
masters
Quote:

Originally Posted by twig684
Cant figure out these 4: Please help.

log2(x + 5) = 4 Find x.

$\displaystyle 2^4=x+5$

Quote:

Originally Posted by twig684

log6(x2 - x) = 1 Find x.

$\displaystyle 6^1=x^2-x$

Quote:

Originally Posted by twig684

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

$\displaystyle \log_4(x(x+6))=2$

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

Quote:

Originally Posted by twig684

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

$\displaystyle \log\frac{x+2}{x}=2$

$\displaystyle 10^2=\frac{x+2}{x}$
• Nov 5th 2008, 01:30 PM
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..?

1. =11
2. =3 or -2
3. =2
4. =2/99
• Nov 5th 2008, 01:34 PM
Moo
Quote:

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 ><
• Nov 5th 2008, 01:41 PM
twig684
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.
• Nov 5th 2008, 01:44 PM
masters
Quote:

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?
• Nov 5th 2008, 01:53 PM
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.
• Nov 5th 2008, 01:58 PM
masters
Quote:

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.

$\displaystyle 16=x^2+6x$

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

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

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

x cannot be negative, so x=2.
• Nov 5th 2008, 02:00 PM
twig684
Thank you very much.