SOLUTION: Kindly explain how to resolve the following logarithmic equation 1) 1 - log (x - 2) = log (3x + 1) Thanks so much!!!

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Question 170375This question is from textbook Precalculus
: Kindly explain how to resolve the following logarithmic equation
1) 1 - log (x - 2) = log (3x + 1)
Thanks so much!!!This question is from textbook Precalculus

Found 2 solutions by Earlsdon, Alan3354:
Answer by Earlsdon(6103) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x:
1-Log%28x-2%29+=+Log%283x%2B1%29 Add Log%28x-2%29 to both sides.
1+=+Log%283x%2B1%29%2BLog%28x-2%29 Apply the "product" rule for logarithms: Log(M)+Log(N) = Log(M*N)
1+=+Log%28%283x%2B1%29%28x-2%29%29 Perform the indicated multiplication.
Log%283x%5E2-5x-2%29+=+1 Recall the definition of the logarithm of a number..."The logarithm of a number is the power to which the base (base = 10 in this case) must be raised to equal that number", so...
10%5E1+=+3x%5E2-5x-2 Subtract 10 from both sides.
3x%5E2-5x-12+=+0 Solve by factoring.
%283x%2B4%29%28x-3%29+=+0 Apply the zero product rule.
3x%2B4+=+0 or x-3+=+0 so...
3x+=+-4 then highlight%28x+=+-4%2F3%29 or
highlight%28x+=+3%29
You can verify these solutions by substituting the values of x into the original equation. Notice that when subsititute x = 3, you'll get real numbers for the logarithms, but when you substitute x = -4/3, you'll get complex numbers but the two sides will be equal in both cases.

Answer by Alan3354(21583) About Me  (Show Source):
You can put this solution on YOUR website!
Kindly explain how to resolve the following logarithmic equation
1) 1 - log (x - 2) = log (3x + 1)
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1 - log(x - 2) = log(3x + 1)
1 = log(3x+1)-log(x-2)
1 = log((3x+1)/(x-2)) Subtracting a log --> dividing by
Raise 10 to both sides
10 = (3x+1)/(x-2)
10*(x-2) = 3x+1
10x-20 = 3x+1
7x = 21
x = 3
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