SOLUTION: Determine the positive root of the following equation: log3(c - 4) + log3(c + 6) = 4. (Please round your answer to one decimal place.) Answer

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Question 453069: Determine the positive root of the following equation:
log3(c - 4) + log3(c + 6) = 4.

(Please round your answer to one decimal place.)

Answer

Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
log3(c - 4) + log3(c + 6) = 4
log3(c - 4)(c + 6) = 4
(c - 4)(c + 6) = 3^4
c^2 + 6c - 4c -24 = 81
c^2 + 2c -105 = 0
apply the quadratic formula to find the roots:
c = {9.3, -11.3}
throw out the negative root leaving:
c = 9.3
.
details of quadratic to follow:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ac%5E2%2Bbc%2Bc=0 (in our case 1c%5E2%2B2c%2B-105+=+0) has the following solutons:

c%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%282%29%5E2-4%2A1%2A-105=424.

Discriminant d=424 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-2%2B-sqrt%28+424+%29%29%2F2%5Ca.

c%5B1%5D+=+%28-%282%29%2Bsqrt%28+424+%29%29%2F2%5C1+=+9.295630140987
c%5B2%5D+=+%28-%282%29-sqrt%28+424+%29%29%2F2%5C1+=+-11.295630140987

Quadratic expression 1c%5E2%2B2c%2B-105 can be factored:
1c%5E2%2B2c%2B-105+=+1%28c-9.295630140987%29%2A%28c--11.295630140987%29
Again, the answer is: 9.295630140987, -11.295630140987. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B2%2Ax%2B-105+%29