How do I condense complex logarithms?

Question

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!

Answer 1

 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

 

     Log_{2 [ ( 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 =

                  

 

 

             

Answer 2

1. Use the relationship log_xa - log_xb = log_x(a/b)

log₂(2x+6) - log₂(x-1) = log₂3

⇒ log₂((2x+6)/(x-1)) = log₂3

 

2. Use the relationship that log_xa = log_xb 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.

Answer 3

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