# Sum of absolute valuesCalculate the sum of the absolute values of solutions of the equation : log(2x+1) (3x+6)=2

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We have to find the sum of the absolute values of the solutions of log(2x+1) (3x+6) = 2.

log(2x+1) (3x+6) = 2

=> 3x + 6 = (2x + 1)^2

=> 3x + 6 = 4x^2 + 1 + 4x

=> 4x^2 + x - 5 = 0

=> 4x^2 + 5x - 4x - 5 = 0

=> x(4x + 5) - 1(4x + 5) = 0

=> (x - 1)(4x + 5) = 0

x = 1 and x = -5/4

The absolute values of the solutions is |1| = 1 and |-5/4| = 5/4

**The sum is 1 + 5/4 = 9/4**

We'll impose the constraints of existence of logarithm:

3x + 6 > 0

3x > -6

x > -2

2x + 1 > 0

x > -1/2

2x + 1 different from 1

x different from 0.

Now, we'll solve taking antilogarithms:

3x+6 = (2x + 1)^2

We'll expand to square:

3x+6 = 4x^2 + 4x + 1

We'll subtract 3x + 1 both sides:

4x^2 + 4x + 1 - 3x - 6 = 0

We'll combine like terms:

4x^2 + x - 5 = 0

We'll apply quadratic formula:

x1 = [-1+sqrt(1 + 80)]/8

x1 = (-1+9)/8

x1 = 1

x2 = (-1-9)/8

x2 = -5/4

Since the second value is not in the interval of admissible values, we'll reject it.

**The only solution is x = 1.**

**The sum of absolute values is:**

**|1| + |-5/4| = 1 + 5/4 = 9/4**