SOLUTION: Log (x^2 +x + 6 )^2 to the base x+1 = 4

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Question 1038580: Log (x^2 +x + 6 )^2 to the base x+1 = 4
Found 2 solutions by robertb, MathTherapy:
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
log%28x%2B1%2C%28x%5E2+%2Bx+%2B+6+%29%5E2%29+=+4
==> %28x%2B1%29%5E4+=+%28x%5E2+%2Bx+%2B+6+%29%5E2, or 2x%5E3-7x%5E2-8x-35+=+0.
The roots of this polynomial are 5, -3%2F4%2B%281%2F4%29%2Asqrt%2847%29%2Ai, and -3%2F4-%281%2F4%29%2Asqrt%2847%29%2Ai. (Go to http://www.mathportal.org/calculators/solving-equations/polynomial-equation-solver.php for a quick solution!)
Hence, x = 5.

Answer by MathTherapy(10557) About Me  (Show Source):
You can put this solution on YOUR website!

Log (x^2 +x + 6 )^2 to the base x+1 = 4
log+%28x+%2B+1%2C+%28x%5E2+%2B+x+%2B+6%29%5E2%29+=+4

%28x+%2B+1%29%5E4+=+%28x%5E2+%2B+x+%2B+6%29%5E2

%28%28x+%2B+1%29%5E2%29%5E2+=+%28x%5E2+%2B+x+%2B+6%29%5E2 ----- Applying a%5E4+=+%28a%5E2%29%5E2

%28x+%2B+1%29%5E2+=+x%5E2+%2B+x+%2B+6 ----------- Equating terms (Exponents are equal and so are the bases)

x%5E2+%2B+2x+%2B+1+=+x%5E2+%2B+x+%2B+6

x%5E2+-+x%5E2+%2B+2x+-+x+=+6+-+1

highlight_green%28x+=+5%29