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# How do I condense complex logarithms?

I'm stuck on a problem: log2(2x+6)-log2(x-1)=log23
I haven't learned how to do it, and I can't seem to find an example or a step-by-step. Please help!

### 3 Answers by Expert Tutors

Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...
4.8 4.8 (4 lesson ratings) (4)
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What is important to realize is that:

Log ( AB) = log A + Log B

Log A/ B = Log A - Log B

Log2 ( 2X +6) - Log 2 ( X - 2) = Log2 3

Log2 [ ( 2X - 6) / X -1 )] = Log2 3

(2X + 6 / X - 1) = 3

3X -3 = 2 X + 6

X = 9

It will be Good Exercise to do the following calculations:

Log6 9 + Log6 4 =

Log3 27 + Log3 3 =

Log 2 8 + Log2 2 =

Jon L. | College Math, Statistics, Finance, Actuarial Science at a Discount!College Math, Statistics, Finance, Actua...
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1. Use the relationship logxa - logxb = logx(a/b)
log2(2x+6) - log2(x-1) = log23
⇒ log2((2x+6)/(x-1)) = log23

2. Use the relationship that logxa = logxb implies a = b
log2((2x+6)/(x-1)) = log23
⇒ (2x+6)/(x-1) = 3

3. Multiply both sides by x-1
(2x+6)/(x-1) = 3
⇒ 2x+6 = 3(x-1)

4. Solve for x
2x+6 = 3(x-1)
⇒ 2x+6 = 3x-3
⇒ 9 = x

So x=9.
Vivian L. | Microsoft Word/Excel/Outlook, essay composition, math; I LOVE TO TEACHMicrosoft Word/Excel/Outlook, essay comp...
3.0 3.0 (1 lesson ratings) (1)
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Hi Jensen;
log2(2x+6)-log2(x-1)=log23
Our first priority is to verify that all logs are to the same base.  These are all to the base of 2.  We may now proceed.
The logarithm-principle we need to apply is...
log a/b=log a - log b
Therefore...
log2[(2x+6)/(x-1)]=log23
We can eliminate log2 on both sides...
(2x+6)/(x-1)=3
(2x+6)/(x-1)=3/1
Let's cross-multiply...
1(2x+6)=3(x-1)
(2x+6)=3(x-1)
2x+6=3x-3
Let's combine like terms...
Let's subtract 2x from both sides...
-2x+2x+6=3x-3-2x
6=x-3
Let's add 3 to both sides...
3+6=x-3+3
9=x

Let's check our work by returning to the original equation...
log2(2x+6)-log2(x-1)=log23
log2[(2*9)+6)-log2(9-1)=log23
log2(18+6)-log2(8)=log23
log2(24)-log2(8)=log23
log a/b=log a - log b
log2[(24)/(8)]=log23
log23=log23