Question 186597: Hi guys, im stumped with this one:
2ln2 - ln(x-1) = ln(2x)
solve for 'x'.
cheers =) Found 2 solutions by nerdybill, josmiceli:Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! You will need to apply "log rules".
Review them at:
http://www.purplemath.com/modules/logrules.htm
.
2ln2 - ln(x-1) = ln(2x)
2ln(2/(x-1)) = ln(2x)
ln(2/(x-1))^2 = ln(2x)
(2/(x-1))^2 = 2x
2^2/(x-1)^2 = 2x
4/(x-1)^2 = 2x
2/(x-1)^2 = x
2 = x(x-1)^2
2 = x(x-1)(x-1)
2 = x(x^2-x-x+1)
2 = x(x^2-2x+1)
2 = x^3-2x^2+x
0 = x^3-2x^2+x-2
Factor (by grouping) on the right:
0 = (x^3-2x^2)+(x-2)
0 = x^2(x-2)+(x-2)
0 = (x-2)(x^2+1)
.
Setting each term (in parenthesis to zero) we get:
x = {2, i}
You can put this solution on YOUR website! In general:
and
check:
OK
The negative root doesn't work since
raising to a power can't
result in a negative
