SOLUTION: solve for x: log(base8)(2x+3) + log(base8)(x+1) = 1

Question 274680: solve for x: log(base8)(2x+3) + log(base8)(x+1) = 1
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
log(base8)(2x+3) + log(base8)(x+1) = 1
.
log(base8)[(2x+3)(x+1)] = 1
.
8^log(base8)[(2x+3)(x+1)] = 8^1
.
(2x+3)(x+1) = 8
.
2x^2+2x+3x+3 = 8
.
2x^2+5x+3 = 8
.
2x^2+5x-5 = 0
.
Since we can't factor, we must resort to the quadratic equation which yields:
x = {0.766, -3.266}
Throw out the negative answer, it is an "extraneous" solution.
leaving:
x = 0.766
.
Details of the quadratic follows:

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=65 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 0.765564437074637, -3.26556443707464. Here's your graph:

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