Question

Solve log2(x+3)=log2(3)-log2(x-2)

Answer 1

`log(x + 3) = log(3) - log(x - 2) \\ log(x + 3) = log( (3)/(x - 2) ) \\ log \: is \: monotonic \:over \: (0 - \infty ) \\ x + 3 = (3)/(x - 2) \\ (x + 3)(x - 2) = 3 \\ x {}^(2) + 3x -2 x - 6 = 3 \\`

`x {}^(2) + x - 9= 0 \\ \delta = 1 - 4(1)( - 9) = 37 \\ x = ( √(37) - 1)/(2) \approx2.54138 \: accepted \\ x = ( - 1 - √(37) )/(2) \approx - 3.54138 \: rejected`

The negative answer is rejected as the input of the logarithmic function is always positive; henceforth we have the following conditions:

`x + 3 > 0 \\ x > - 3 \\ \\ x - 2 > 0 \\ x > 2`

Answer 2

We have the conditions : x + 3 > 0 and x - 2 > 0 ;
log2 ( x + 3 ) = log2 [ 3 / ( x - 2 ) ] ;
x + 3 = 3 / ( x - 2 ) ;
( x + 3 )( x - 2 ) = 3 ;
x^2 + 3x - 2x - 6 = 3 ;
x^2 + x - 9 = 0 ;
For ax^2 + bx + c = 0, the value of x is given by: ;
a = 1 ; b = 1 ; c = - 9 ;
then, x 1 ≈ 2.54 ; x 2 ≈ - 3.54 ;
The solutions are 2.54 and - 3.54 because its verify the conditions.