log(2x) + log(x+5) = 2

First, we'll discuss the constraints of existence of logarithms:

2x>0 and x+5>0

x>-5

x>0

So, for both logarithms to exist, the values of x have to be in the interval (0,+inf.).

Now, we'll solve the equation, using the product property of logarithms: the sum of logarithms is the logarithm of the product.

log(2x) + log(x+5) = log [2x(x+5)]

The equation will become:

log [2x(x+5)] = 2

But 1  = log 10

We'll re-write the equation:

log [2x(x+5)] = 2 log 10

log [2x(x+5)] = log 100

We'll use the one to one property:

[2x(x+5)]  = 100

We'll open the brackets:

2x^2 + 10x - 100 = 0

We'll divide by 2:

x^2 + 5x - 50 = 0

We'll apply the quadratic formula:

x1 = [-5+sqrt(25+200)]/2

x1 = (-5+15)/2

x1 = 5

x2 = (-5-15)/2

x2 = -20/2

x2 = -10

Since the second value of x is not in the interval of convenient values, the equation will have only a solution, namely x = 5.