Can someone please check my answer to this logarithmic equation?
Here is the question:
log(base)2 (x + 1) + log(base)2 (x - 1) = log(base)2 8
My answer: x = 4.08 or -2.08

Question

Can someone please check my answer to this logarithmic equation?
Here is the question: 
log(base)2 (x + 1) + log(base)2 (x - 1) = log(base)2 8
My answer: x = 4.08 or -2.08

sighn0more · Answer

And if I got the answer wrong, what do I need to do to solve the problem correctly?

nnesha · Answer

work > answer 
most important thin HOW  did you get the answer 
not what is the answer :=))
so please explain how did you get that :=))

nnesha · Answer

thing*

sighn0more · Answer

Okay! So I used the rules of logs to get: logbase2 (x + 1)(x-1) = logbase2 8, and then I expanded the binomial to get x^2 - 2x - 1 = logbase2 8. I then subtracted 8 from both sides, getting x^2 - 2x - 9 = 0, and then I used the quadratic equation to get my answer.

nnesha · Answer

(x+1)(x-1) = ??
hint: difference of squares (a+b)(a-b) =a^2 -b^2
both parentheses are same but with opposite sign you can apply difference of squares

sighn0more · Answer

OH, I see. I completely forgot about that property. Okay, so it would be x^2 - 1^2

nnesha · Answer

right which is same as x^2 -1 :=))

sighn0more · Answer

Awesome! Would I then subtract 8 from both sides of the equation?

nnesha · Answer

yes right

sighn0more · Answer

Okay, cool! My new answer is x = 19.

nnesha · Answer

hmm how ?

sighn0more · Answer

I put x^2 - 9 into the quadrtic equation >  $$\sqrt{38}$$ / 2
and got 19

anonymous · Answer

i wouldn't subtract 8  from both sides, i would add 1 to both sides and solve $$x^2=9$$ in my head

nnesha · Answer

well difference of squares 
x^2 -9=  ??

anonymous · Answer

think of a number whose square is 9

anonymous · Answer

bet you get it on the first try

sighn0more · Answer

Ah! So it would be 3.

sighn0more · Answer

Or would it be both +3 and -3...

nnesha · Answer

right $$x^2-9 =(x+3)(x-3)$$
solve both binomial for x 
x+3 = 0   and x-3=0
x=-3   x=3 
-3 is an extraneous solution so that wouldn't be the asnwer

nnesha · Answer

when u substitute x for - 3 you will not get the equal sides 
that means -3 isn't a solution  $$\log_2 (-3+1) +\log_2 (-3-1) \cancel{=} \log_2 8$$

nnesha · Answer

so -3 isn't a solution

nnesha · Answer

make sense ?? :=))

sighn0more · Answer

Yes! Perfect sense. Wow, that took a lot of steps; thanks for hanging in there with me!

nnesha · Answer

np \o^_^o/ good work keep it up!