Saturday, December 14, 2013

log x + log(2x-8) =1

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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