|
Question 166489: log (x-3)+ log(x) = 1 and log2 (x2-6x)=3+log2(1-x)
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! log (x-3)+ log(x) = 1
applying log rule:log(mn) = log(m) + log(n)
log [(x-3)(x)] = 1
(x-3)(x) = 10^1
x^2-3x = 10
x^2-3x-10=0
(x-5)(x+2) = 0
x = {-2, 5}
.
.
.
log2(x^2-6x)=3+log2(1-x)
log2(x^2-6x) - log2(1-x) = 3
applying log rule:log(m/n) = log(m) – log(n)
log2[(x^2-6x)/(1-x)] = 3
(x^2-6x)/(1-x) = 2^3
(x^2-6x)/(1-x) = 8
x^2-6x = 8(1-x)
x^2-6x = 8-8x
x^2+2x-8 = 0
(x+4)(x-2) = 0
x = {-4, 2}
|
|
|
| |
