SOLUTION: log(Base3)(x-1) -Log(base3)(x+2)=2

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Question 79779: log(Base3)(x-1) -Log(base3)(x+2)=2
Found 2 solutions by rapaljer, MathTherapy:
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
log%283%2C%28x-1%29%29+-log%283%2C%28x%2B2%29%29=2
log%283%2C%28%28x-1%29%2F%28x%2B2%29%29%29+=2

3%5E2=%28x-1%29%2F%28x%2B2%29+
9=%28x-1%29%2F%28x%2B2%29+

Multiply both sides by the common denominator : %28x%2B2%29
9%28x%2B2%29+=+x-1
9x%2B18=x-1
9x-x%2B18=-1
8x=-1-18
8x=-19
x=-19%2F8

However, there is a restriction that you can't have a log of a negative. This answer for x=-19%2F8 gives the log of a negative number in BOTH logarithmic expressions, so it must be rejected. Therefore there is NO SOLUTION to this problem.

For additional exercises and detailed solution IN COLOR, see my webpage on logarithms by clicking on my tutor name "rapaljer" anywhere in algebra.com. Then look for "Math in Living Color", "College Algebra", "Section 4.04 Solving Exponential and Logarithmic Equations."

R^2 at SCC

Answer by MathTherapy(10699) About Me  (Show Source):
You can put this solution on YOUR website!
log(Base3)(x-1) -Log(base3)(x+2)=2

log+%283%2C+%28x+-+1%29%29+-+log+%283%2C+%28x+%2B+2%29%29+=+2
For log (a), "a" MUST be > 0
So, we set the SMALLER "a" (x - 1) > 0____x > 1

We now get: log+%283%2C+%28x+-+1%29%29+-+log+%283%2C+%28x+%2B+2%29%29+=+2, with x > 1
log+%283%2C+%28%28x+-+1%29%2F%28x+%2B+2%29%29%29+=+2 --- Applying log+%28a%2C+%28b%29%29+-+log+%28a%2C+%28c%29%29 = log+%28a%2C+%28b%2Fc%29%29
      %28x+-+1%29%2F%28x+%2B+2%29+=+3%5E2 -- Converting to EXPONENTIAL form 
   9(x + 2) = x - 1 ---- Cross-multiplying
    9x + 18 = x - 1
     9x - x = - 1 - 18
         8x = - 19

As seen above, x is NEGATIVE (< 0), so x is NOT > 1. As a result, NO SOLUTIONS exist here!