Logarithms?

Question

Hey,

I've got a que about this:
log(sub)2 (-x - 3) - log(sub)2 (x - 1) - log(sub)2(x + 3) = 1

I did this first -

log(sub)2 [(-x - 3) / (x - 1) / (x + 3)] = 1

and then i just cant figure out.
All suggestions greatly appriciated.

Answer 1

You may use this:

log(sub)2 [(-x - 3) / (x - 1) / (x + 3)] = 1
2^[log(sub)2 [(-x - 3) / (x - 1) / (x + 3)] ]= 2^[1]
(-x - 3) /[(x - 1)(x + 3)] = 2
(-x-3) = 2(x - 1)(x + 3)

The rest it's all yours...

Answer 2

In the following,"log" means log to base 2:-
log(-x - 3) - log(x - 1) - log(x + 3) = 1
(-x - 3) / ((x - 1).(x + 3)) = 2
(-x - 3) = 2.(x - 1).(x + 3)
(-x - 3) = 2.(x² + 2x - 3)
2x² + 5x - 3 = 0
(2x - 1).(x + 3) = 0
x = 1/2 , x = - 3

Answer 3

It is really log(sub)2[(-x-3)/[(x-1)(x+3)] =1

reduce; log(sub)2[-1/(x-1)] =1
2^1 = -1/(x-1)
2(x-1) = -1
2x - 2= -1
2x = 1
x = 1/2

Answer 4

OK that looks nearly correct. What you need is

log(sub)2[(-x-3)/((x-1)*(x+3))

Add logarithms to multiply numbers.

Answer 5

You got a test tomorrow too,sorry i cant help,I only got 30percent in my last one.