SOLUTION: Solve. Log (〖3x〗^2 + 2x +2)=0.84509

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Question 202643: Solve. Log (〖3x〗^2 + 2x +2)=0.84509
Answer by jsmallt9(3759) About Me  (Show Source):
You can put this solution on YOUR website!
If the problem is not: log%28%283x%5E2+%2B2x+%2B+2%29%29+=+0.84509, please repost it
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If the problem is log%28%283x%5E2+%2B2x+%2B+2%29%29+=+0.84509, start by rewriting it in exponential form:
3x%5E2+%2B+2x+%2B+2+=+10%5E%280.84509%29
10%5E%280.84509%29+=+6.9998704114807013 Since this is so close to 7 I think you are supposed to use 7. (If not then you will have to use the quadratic formula on the equation with this very long decimal!?)
3x%5E2+%2B+2x+%2B+2+=+7
Subtracting 7 from both sides:
3x%5E2+%2B+2x+-+5+=+0
Factoring:
%283x+%2B+5%29%28x+-+1%29+=+0
The only way for a product to be zero is if one of the factors is zero. So:
%283x+%2B+5%29+=+0 or %28x+-+1%29+=+0
Solving these we get:
x+=+%28-5%29%2F3 or x+=+1