SOLUTION: I have a question about the logarithmic equations. I'll use log[a, b] where a is the base of the logarithm. Is log[sqrt3,(1-x)^2] - log[sqrt3, (3-x)] < 2 equal to 2 *

Algebra ->  Inequalities -> SOLUTION: I have a question about the logarithmic equations. I'll use log[a, b] where a is the base of the logarithm. Is log[sqrt3,(1-x)^2] - log[sqrt3, (3-x)] < 2 equal to 2 *      Log On


   

Question 996322: I have a question about the logarithmic equations. I'll use log[a, b] where a is the base of the logarithm.

Is
log[sqrt3,(1-x)^2] - log[sqrt3, (3-x)] < 2
equal to
2 * log[sqrt3, (1-x)] - log [sqrt3, (3-x)]<2
If they're not equal, why aren't they? Isn't the logarithm property of exponent is log(1-x)^2 = 2 * log(1-x)?
Found 2 solutions by MathLover1, rothauserc:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

log%28sqrt%283%29%2C%281-x%29%5E2%29+-+log%28sqrt%283%29%2C+%283-x%29%29+%3C+2

log%28sqrt%283%29%2C%281-x%29%5E2%2F+%283-x%29%29+%3C+2...........change to base 10
log%28%281-x%29%5E2%2F+%283-x%29%29%2Flog%28sqrt%283%29%29+%3C+2
log%28%281-x%29%5E2%2F+%283-x%29%29+%3C+2log%28sqrt%283%29%29
log%28%281-x%29%5E2%2F+%283-x%29%29+%3C+log%28%28sqrt%283%29%29%5E2%29
log%28%281-x%29%5E2%2F+%283-x%29%29+%3C+log%283%29....if log same, then
%281-x%29%5E2%2F+%283-x%29+%3C+3......solve for x
%281-x%29%5E2+%3C+3%283-x%29
1-2x%2Bx%5E2+%3C+9-3x
1-2x%2Bx%5E2-9%2B3x%3C0
x%5E2%2Bx-8%3C0.....use quadratic formula
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-1+%2B-+sqrt%28+1%5E2-4%2A1%2A%28-8%29+%29%29%2F%282%2A1%29+
x+=+%28-1+%2B-+sqrt%28+1%2B32+%29%29%2F2+
x+=+%28-1+%2B-+sqrt%28+33+%29%29%2F2+
exact solutions:
x+=+%281%2F2%29%28-1+%2B+sqrt%28+33+%29%29+
x+=+%281%2F2%29%28-1+-+sqrt%28+33+%29%29+


Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
use the division logarithm rule
log[sqrt3,(1-x)^2] - log[sqrt3, (3-x)] = log[sqrt3, (1-x)^2 / (3-x)], therefore
log[sqrt3, (1-x)^2 / (3-x)] < 2