SOLUTION: log(base 3)(x+8) + log(base 3)(x) = 2 I'm not for sure if I am sitting this up right. So far I have: log(base 3)( x+8/x ) = 2 Should I multiply or divide since its a +?

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Question 165215This question is from textbook
: log(base 3)(x+8) + log(base 3)(x) = 2
I'm not for sure if I am sitting this up right.
So far I have:
log(base 3)( x+8/x ) = 2 Should I multiply or divide since its a +?
Any help appreciated.
This question is from textbook

Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website!
For log rules see:
http://www.purplemath.com/modules/logrules.htm
.
log(base 3)(x+8) + log(base 3)(x) = 2
.
Applying log rule: logb(mn) = logb(m) + logb(n)
log(base 3)(x^2+8x) = 2
.
Applying log rule: logb(m) = n =>> m = b^n
x^2+8x = 3^2
.
x^2+8x = 8
x^2+8x-8 = 0
.
Since we can't factor, use the quadratic equation. Doing so will yield:
x = {0.899, -8.899}
.
Details of quadratic:

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=96 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 0.898979485566356, -8.89897948556636. Here's your graph:

= 0