SOLUTION: How do I solve the following logarithm for x? The base of the following logs is 2. log(x-1) - log(x+3) = log(1/x)

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: How do I solve the following logarithm for x? The base of the following logs is 2. log(x-1) - log(x+3) = log(1/x)      Log On


   

Question 698510: How do I solve the following logarithm for x?
The base of the following logs is 2.
log(x-1) - log(x+3) = log(1/x)
Found 2 solutions by stanbon, MathLover1:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The base of the following logs is 2.
log2(x-1) - log2(x+3) = log2(1/x)
-----
log2[(x-1)/(x+3)] = log2(1/x)
------
Take the antilog of both sides:
(x-1)/(x+3) = 1/x
----
Cross-multiply:
x^2-x = x+3
------
x^2 -2x - 3 = 0
---
Factor:
(x-3)(x+1) = 0
Positive solution:
x = 3
============
Cheers,
Stan H.
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Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

log%28x-1%29+-+log%28x%2B3%29+=+log%281%2Fx%29

log%28%28x-1%29%2F%28x%2B3%29%29+=+log%281%2Fx%29
%28x-1%29%2F%28x%2B3%29+=+1%2Fx
%28x-1%29x+=+1%28x%2B3%29
x%5E2-x+=+x%2B3
x%5E2-x-x-3=0
x%5E2-2x-3=0
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-%28-2%29+%2B-+sqrt%28%28-2%29%5E2-4%2A1%2A%28-3%29+%29%29%2F%282%2A1%29+
x+=+%282+%2B-+sqrt%284%2B12+%29%29%2F2+
x+=+%282+%2B-+sqrt%2816+%29%29%2F2+
x+=+%282+%2B-+4%29%2F2+
solutions:
x+=+%282+%2B+4%29%2F2+
x+=+6%2F2+
x+=+3+
or
x+=+%282-+4%29%2F2+
x+=+-2%2F2+
x+=-1+