solve for x... log2(x^2-1)-log2(x+1)=3?

Use the properties of logarithms:

i.e, that log(a) - log(b) = log(a/b)

Then log2(x^2 -1) - log2(x+1) = 3 implies

log2(x^2-1)/(x+1) = 3

Then 2^3 = (x^2-1)/(x+1)

8 = (x+1)(x-1)/(x+1)

Then 8 = x-1 as the (x+1)'s cancel on the right hand side

==> Solution is x = 9

(a million/3)^2 - x = 27 -x = 27 - (a million/3)^2 x = (a million/3)^2 - 27 in line with hazard you meant (a million/3)^(2 - x) = 27. consequently, (a million/3)^(2 - x) = 27 (a million/3)^(2 - x) = (a million/3)^(-3) 2 - x = -3 -x = -5 x = 5

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

.x^2-1=8(x+1)
x^2-1=8x+8
x^2-8x-1-8=0
x^2-8x-9=0
(x+1)(x-9)=0
x+1=0 x-9=0
x=-1 x=9

x=-1,9 answer//

x-1=8
x=9
God bless you.

x=9

log_2(x^2 - 1) - log_2(x + 1) = 3
log_2[(x^2 - 1^2)/(x + 1)] = 3
log_2{[(x + 1)(x - 1)]/(x + 1)} = 3
log_2(x - 1) = 3
x - 1 = 2^3
x = 8 + 1
x = 9

log2((x^2-1)/(x+1)=3
x-1=8
x=9