Question

log2(x+1) + log2(x + 3) = 3
Find the value for x

Answer 1

log2 (x+1) +log2 (x+3)=3

=> log2 (x+1)(x+3)=3

=> log2 (x^2+4x+3)=3

=>(x^2+4x+3)=2^3

=> x^2+4x+3-8=0

=> x^2+4x-5=0

=> x^2+5x-x-5=0

=>x (x+5)-1 (x+5)=0

=>(x-1)(x+5)=0

x=1,-5

but log to be defined
x+1> 0 or, x>-1
and x+3> 0 or x>-3

take common x>-1 and x>-3
e.g. x>-3

hence x always greater then -3
so, -5 doesn't possible
hence x=1 is only answer

Answer 2

log 2 (x + 1) + log 2 (x + 3) = 3
=>log 2 (x + 1)(x + 3) = 3              (Using property log a mn = log a m + log a n)
=> log 2 ( x² + x + 3x + 3) = 3
=>log 2 ( x² + 4x + 3) = 3
=> x² + 4x + 3 = 2³ =8
=> x² + 5x - x -5 = 8
=> x(x + 5) - 1( x +5) -5 = 0
=> (x - 1)( x + 5) = 0

=> x - 1 = 0
=> x = 1
     OR
=> x + 5 = 0
=> x = -5

Now putting the values of x we get,
log 2 ( -5 + 1) + log 2 ( -5 + 3) = log 2 -4 + log 2 -2
But -4 and -2 are not defined

So x = 1