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.

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- Nov 5th 2008, 02:05 PMtwig684Logarithmic 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, 02:25 PMmasters
- Nov 5th 2008, 02:30 PMtwig684
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, 02:34 PMMooQuote:

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..?

Remember, in a test you won't have the answers, and you'll need the method to solve them >< - Nov 5th 2008, 02:41 PMtwig684
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, 02:44 PMmasters
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, 02:53 PMtwig684
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, 02:58 PMmasters
- Nov 5th 2008, 03:00 PMtwig684
Thank you very much.