SOLUTION: solve log (x+2) + log (2x-3) =2 x>0

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Question 864704: solve
log (x+2) + log (2x-3) =2 x>0
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

log (x+2) + log (2x-3) =2 

log (x+2)(2x-3) = 2

(x+2)(2x-3) = 100

 2x^2 +x -100 = 0

x = 6.82548584904245    x > 0

 


 
Solved by pluggable solver: SOLVE quadratic equation with variable


Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B1x%2B-100+=+0) has the following solutons:

  

  x%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=%281%29%5E2-4%2A2%2A-100=801.

  

  Discriminant d=801 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+801+%29%29%2F2%5Ca.

  

    x%5B1%5D+=+%28-%281%29%2Bsqrt%28+801+%29%29%2F2%5C2+=+6.82548584904245

    x%5B2%5D+=+%28-%281%29-sqrt%28+801+%29%29%2F2%5C2+=+-7.32548584904245

    

    Quadratic expression 2x%5E2%2B1x%2B-100 can be factored:

  2x%5E2%2B1x%2B-100+=+2%28x-6.82548584904245%29%2A%28x--7.32548584904245%29

  Again, the answer is: 6.82548584904245, -7.32548584904245.
Here's your graph:

graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B1%2Ax%2B-100+%29