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

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log2 (x-1) = 2 -log2(x-2)

Solution:

We rewrite the equation converting 2 as log 2 base 2 = log 2 (2^2)

log 2 (x-1) = log2 (2^2) - log2 (x+2)

log(2x-1) = log2 {2^2 / (x+2)}, as log a- log b = log (a/b)Take antilog.

x-1 = 2^2 /(x+2 = 4/(x+2). Multiply by 4.

(x-1)(x+2) = 4

x^2+x-2 -4 = 0

x^2+x-6 = 0

(x+3)(x-2) = 0

x+3 = 0, x-2 = 0

x=-3, x= 2

We'll move log 2 (x+2) to the left side and we'll get:

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

We'll use the product property of logarithms:

log 2 [(x-1)(x+2)] = 2

[(x-1)(x+2)] = 2^2

We'll remove the brackets:

x^2 + 2x - x - 2 = 4

x^2 + x - 2 - 4 = 0

x^2 + x - 6 = 0

We'll apply the quadratic formula:

x1 = [-1+(1+24)]/2

x1 = (-1+5)/2

x1 = 2

x2 = (-1-5)/2

x2 = -3

From the contraints of existance of logarithms, we'll get:

x-1>0

x>1

x+2>0

x>-2

So, in order to respect both conditions, x has to have values in the interval (1,+inf.).

Because the second solution, x2 = -3, does not belong to the interval (1,+inf.), the equation will have only one solution, namely:

**x = 2**