Math Question. Logerithims?

Question

Solve:
Log[2] (X+4) + Log[2] (X) = 5

Answer 1

32=2^5=2^(Log[2] (X+4) + Log[2] (X))
=2^(Log[2] (X+4))*2^(Log[2] (X))
=(X+4)*X

So (X+4)*X=32.
Solving, we get
X=4 or -8.
Substituting X=-8 in the original equation fails, but X=4 satisfies it.

Answer 2

In the following, log is taken to mean log to base 2.
log(x + 4) + log x = 5
log [ (x).(x + 4) ] = 5
(x).(x + 4) = 2^5
x² + 4x - 32 = 0
(x + 8).(x - 4) = 0
x = - 8 , x = 4
Accept only the +ve value for x:-
Answer x = 4

Answer 3

Log[2] (X+4) + Log[2] (X) = 5
Log[2] (X+4)X=5
2^Log[2] (X+4)X=2^5
X^2+4X=32
X^2+4X-32=0
(-4+-12)/2 X=4 et X=-8(but X must be>0)
So X=4

Answer 4

Law 1 allows us to combine the first 2 expressions into:

log[2] (x(x+4) = 5

So

2^5 = x^2 + 4x

x^2 + 4x - 32 = 0

(x+8)(x-4) = 0

x = -8

x = 4

Answer 5

Using the product rule
Log[2] (x(x+4)) = 5
so x^2 +4x = 2^5
x^2 +4x - 32 = 0
solving the quadratic equation we get
x = -8 or x = 4

Answer 6

when your adding logs, you can drop the log if its the same value. x+4+x=5
2x=1
x=1/2 ?
Log[2]=1/2 ?
im not sure though