SOLUTION: how do you solve log x + log (3x + 5)= 1

Question 745948: how do you solve log x + log (3x + 5)= 1
Found 2 solutions by savvyhush23, nerdybill:
Answer by savvyhush23(50) (Show Source): You can put this solution on YOUR website!
how do you solve log x + log (3x + 5)= 1
Properties of Logarithm:
log M + log N = log MN
log (10) = 1
So,

eliminate log

,it is a quadratic equation,

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=145 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 1.17359909646538, -2.84026576313205. Here's your graph:

Therefore, x = {1.1736,-2.8403}
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
log x + log (3x + 5)= 1
log x(3x + 5)= 1
x(3x + 5)= 10^1
x(3x + 5)= 10
3x^2 + 5x= 10
3x^2 + 5x - 10 = 0
Applying the "quadratic formula" yields:
x = {1.17, -2.84}
throw out the negative solution (extraneous) leaving:
x = 1.17
.
Details of "quadratic formula" 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=145 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 1.17359909646538, -2.84026576313205. Here's your graph:

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