_{2}(2x+6)-log

_{2}(x-1)=log

_{2}3

I'm stuck on a problem: log_{2}(2x+6)-log_{2}(x-1)=log_{2}3

I haven't learned how to do it, and I can't seem to find an example or a step-by-step. Please help!

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Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...

What is important to realize is that:

Log ( AB) = log A + Log B

Log A/ B = Log A - Log B

Log_{2} ( 2X +6) - Log _{2 }( X - 2) = Log_{2} 3

Log_{2 [ ( 2X - 6) / X -1 )] = }Log_{2} 3

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

3X -3 = 2 X + 6

X = 9

It will be Good Exercise to do the following calculations:

Log_{6} 9 + Log_{6} 4 =

Log_{3} 27 + Log_{3} 3 =

Log _{2 }8 + Log_{2} 2 =

1. Use the relationship log_{x}a - log_{x}b = log_{x}(a/b)

log_{2}(2x+6) - log_{2}(x-1) = log_{2}3

⇒ log_{2}((2x+6)/(x-1)) = log_{2}3

2. Use the relationship that log_{x}a = log_{x}b implies a = b

log_{2}((2x+6)/(x-1)) = log_{2}3

⇒ (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.

Hi Jensen;

log_{2}(2x+6)-log_{2}(x-1)=log_{2}3

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...

log_{2}[(2x+6)/(x-1)]=log_{2}3

We can eliminate log_{2} 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

Let's check our work by returning to the original equation...

log_{2}(2x+6)-log_{2}(x-1)=log_{2}3

log_{2}[(2*9)+6)-log_{2}(9-1)=log_{2}3

log_{2}(18+6)-log_{2}(8)=log_{2}3

log_{2}(24)-log_{2}(8)=log_{2}3

log a/b=log a - log b

log_{2}[(24)/(8)]=log_{2}3

log_{2}3=log_{2}3

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