xWhat is x if lg(2x+2)=lg(8+3x)/(x-2)?
We have to solve lg(2x+2)=lg[(8+3x)/(x-2)]
Equate 2x + 2 = (8+3x)/(x-2)
=> (2x + 2)(x - 2) = 8 + 3x
=> 2x^2 + 2x - 4x - 4 = 8 + 3x
=> 2x^2 - 5x - 12 = 0
=> 2x^2 - 8x + 3x - 12 = 0
=> 2x(x - 4) + 3(x - 4) = 0
=> (2x + 3) = 0 and (x - 4) = 0
=> x = -3/2 and x = 4.
For x = -3/2, the given logarithms become logs of negative numbers and are undefined.
The solution of the equation is x = 4
We'll impose the constraints of existence of logarithms:
The interval of admissible values for x is (2, +infinite).
Now, we'll solve the equation:
Since the logarithms have the matching bases, we'll apply the one to one rule:
2x + 2 = [(8+3x)/(x-2)]
We'll cross multiply and we'll get:
[(2x+2)*(x-2)] = 8 + 3x
We'll remove the brackets from the left side:
2x^2 - 4x + 2x - 4 = 8 + 3x
We'll move all terms to the left side and we'll use symmetric property:
2x^2 - 4x + 2x - 4 - 3x - 8 = 0
2x^2 - 5x - 12 = 0
We'll apply quadratic formula:
x1 = [5 + sqrt(25 + 96)]/4
x1 = (5+11)/4
x1 = 16/4
x1 = 4
x2 = -6/4
x2 = -3/2
Since the second solution does not belong to the range of admissible values, we'll reject it.
The only admissible solution is x = 4.