SOLUTION: log (x-3)+ log(x) = 1 and log2 (x2-6x)=3+log2(1-x)
Algebra.Com
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}
RELATED QUESTIONS
Log X + log(6x+1)=log2
(answered by Boreal)
(a) (4x+2)(2x) = 32
(b) (3x+2)(9x+3) = 81
(c) log (x-3) + log (x) = 1
(d) log2 (answered by jim_thompson5910)
solve :... (answered by rapaljer)
logx+log(x+1)=log2
(answered by josgarithmetic,ikleyn)
log10x-log(x-3)=log2 (answered by MathLover1)
log(4+x)-log(x-3)=log2 (answered by farah)
Log[log(2+log2(x+1))]=0 (answered by Alan3354)
How would I solve this?
log2(x-3)-log2(2x+1)=log24 (log base 2,... (answered by jsmallt9)
log2(x-1)+log2(x+10=3 (answered by nerdybill)