First, we'll discuss the constraints of existence of logarithms:
x>0 and 2x-8>0
2x>8
x>4
So, for both logarithms to exist, the values of x have to be in the interval (4,+inf.).
Now, we'll solve the equation, using the product property of logarithms: the sum of logarithms is the logarithm of the product.
log x + log(2x-8) = log [x(2x-8)]
The equation will become:
log [x(2x-8)] = 1
But 1 = log 10
We'll re-write the equation:
log [x(2x-8)] = log 10
We'll use the one to one property:
[x(2x-8)] = 10
We'll opent the brackets:
2x^2 - 8x - 10 = 0
We'll divide by 2:
x^2 - 4x - 5 = 0
We'll apply the quadratic formula:
x1 = [4+sqrt(16+20)]/2
x1 = (4+6)/2
x1 = 5
x2 = (4-6)/2
x2 = -2/2
x2 = -1
Since the second value of x is not in the interval of convenient values, the equation will have only a solution, namely x = 5.
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