SOLUTION: log(x-3)+log(x) = 2 + log(x+2) *The base for the logarithms is 2. I just need help clarifying my answer, I found that this problem had no solution but I'm not completely sure if th

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: log(x-3)+log(x) = 2 + log(x+2) *The base for the logarithms is 2. I just need help clarifying my answer, I found that this problem had no solution but I'm not completely sure if th      Log On


   

Question 694069: log(x-3)+log(x) = 2 + log(x+2) *The base for the logarithms is 2. I just need help clarifying my answer, I found that this problem had no solution but I'm not completely sure if that is correct
Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
log(x-3)+log(x) = 2 + log(x+2)
---------------------------
log[(x-3)x] - log(x+2) = 2
-----
log[(x^2-3x)/(x+2)] = 2
----
(x^2-3x)/(x+2) = 2^2
-------
x^2-3x = 4x+8
x^2 - 7x - 8 = 0
Factor:
(x-8)(x+1) = 0
Positive solution:
x = 8
====================
Cheers,
Stan H.
===================

Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
No, it has a solution.

The four rules of logarithms are:

1.  log%28b%2C%28A%29%29%2Blog%28b%2C%28C%29%29=log%28b%2C%28AC%29%29

2.  log%28b%2C%28A%29%29-log%28b%2C%28C%29%29=log%28b%2C%28A%2FC%29%29

3.  log%28b%2C%28A%2AC%29%29=C%2Alog%28b%2C%28A%29%29

4.  log equation log%28b%2C%28A%29%29=C is equivalent to
exponential equation A=b%5EC

5.  log%28b%2C%28n%29%29=1



       log2(x-3) + log2(x) = 2 + log2(x+2)

Use rule 1 on the left side:

              log2[(x-3)x] = 2 + log2(x+2)

Distribute

               log2(x²-3x) = 2 + log2(x+2)

Get bothlogs on the left side of the equation:

   log2(x²-3x) - log2(x+2) = 2

Use rule 2 on the left

              log2%28%28x%5E2-3x%29%2F%28x%2B2%29%29 = 2

Use rule 4

               %28x%5E2-3x%29%2F%28x%2B2%29 = 22

Write 2² as 4

              %28x%5E2-3x%29%2F%28x%2B2%29 = 4

Multiply both sides by x+2

              x² - 3x = 4(x + 2)

              x² - 3x = 4x + 8

          x² - 7x - 8 = 0

       (x - 8)(x + 1) = 0

       x - 8 = 0;  x + 1 = 0
           x = 8;      x = -1

The only solution is x = 8.  Logarithms of negative
numbers are not real numbers.  So we discard the
negative answer.

Checking:

       log2(x-3) + log2(x) = 2 + log2(x+2)
       log2(8-3) + log2(8) = 2 + log2(8+2)
        log2(5) + log2(2³) = 2 + log2(10)

Use rule 3 on 2nd term, Write 10 as 2·5

       log2(5) + 3·log2(2) = 2 + log2(2·5)

Use rule 5 to rewrite log2(2) as 1
Use rule 1 to rewrite last term on right

       log2(5) + 3·1 = 2 + log2(2) + log2(5)

Use rule 5 to rewrite log2(2) as 1

         log2(5) + 3 = 2 + 1 + log2(5)

Add 2 + 1

         log2(5) + 3 = 3 + log2(5)

So it checks.

Edwin