# solve the eqation: log (2x+2) = 1 - log 3x

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We'll impose the constraints of existence of logarithms:

2x+ 2 >0

2x>-2

x>-1

3x>0

x>0

The interval of admissible solutions for the given equation is: (0 ; +infinite).

Now, we'll sove the equation, adding log 3x both sides:

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

We'll apply the product rule:

log [6x(x+1)] = 1

We'll take antilogarithm:

6x(x+1) = 10

We'll remove the brackets:

6x^2 + 6x - 10 = 0

We'll divide by 2:

3x^2 + 3x - 5 = 0

We'll apply the quadratic formula:

x1 = [-3+sqrt(9 + 60)]/6

x1 = [-3+sqrt(69)]/6

x2 = [-3-sqrt(69)]/6

Since the 2nd value is negative, it doesn't belong to the range of admissible values, so we'll reject it.

**The only valid solution of the equation is: **

**x1 = [-3+sqrt(69)]/6**

To solve the equation: log (2x+2) = 1 - log 3x

We rewrite equation as below by adding log3x to both sides:

log(2x+2)+log3x = 1

log (2x+2)*3x = 1.

=> (2x+2)*3x = 10^1 , As log a = b => a = 10^b by definition.

=> 6x^2+6x-10 = 0.

=> 3x^2+3x -5 = 0.

We use quadratic formula :

x1 = (-3+sqrt(3^2+4*3*5)}/2*3 = (-+sqrt69)/6 .

So x1 = (-3+69^(1/2))/6

x2= (-3+(69)^(1/2))/6.