SOLUTION: log6 (sqrt x) + log6 (sqrt 6x+5) = 1

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: log6 (sqrt x) + log6 (sqrt 6x+5) = 1      Log On


   

Question 206898This question is from textbook Intermediate Algebra
: log6 (sqrt x) + log6 (sqrt 6x+5) = 1 This question is from textbook Intermediate Algebra

Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
If this is whagt you meant:
log%286%2Csqrt%28x%29%29+%2B+log%286%2Csqrt+%286x%2B5%29%29+=+1
.
Then, applying "log rules" we get:
log%286%2Csqrt%28x%29sqrt%286x%2B5%29%29+=+1
log%286%2Csqrt%286x%5E2%2B5x%29%29+=+1
sqrt%286x%5E2%2B5x%29+=+6%5E1
sqrt%286x%5E2%2B5x%29+=+6
6x%5E2%2B5x+=+36
6x%5E2%2B5x-36+=+0
6x%5E2%2B5x-36+=+0
Solving the above with the quadratic equation yields the following:
x = {2.068, -2.901}
We can toss out the negative solution leaving:
x = 2.068
.
Details of quadratic follows:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 6x%5E2%2B5x%2B-36+=+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=%285%29%5E2-4%2A6%2A-36=889.

Discriminant d=889 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-5%2B-sqrt%28+889+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%285%29%2Bsqrt%28+889+%29%29%2F2%5C6+=+2.06800858597926
x%5B2%5D+=+%28-%285%29-sqrt%28+889+%29%29%2F2%5C6+=+-2.9013419193126

Quadratic expression 6x%5E2%2B5x%2B-36 can be factored:
6x%5E2%2B5x%2B-36+=+6%28x-2.06800858597926%29%2A%28x--2.9013419193126%29
Again, the answer is: 2.06800858597926, -2.9013419193126. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+6%2Ax%5E2%2B5%2Ax%2B-36+%29