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November 27, 2014

November 27, 2014

Posted by **alicia** on Wednesday, August 6, 2008 at 10:42pm.

log base 3 of x^2= 2log base 3 of 4-4log base 3 of 5

Solve each of the inequalities:

(a) x^2+2x-3<(or equal to)0

(b) (2x-1)/(3x-2)<(or equal to)1

Any help is appreciated!

- math (1 of 3) -
**drwls**, Thursday, August 7, 2008 at 2:40amlog base(3)x^2 = 2log(3)4 - 4log(3)5

= log(3)[4^2/5^4]

Therefore x^2 = 16/625

x = 4/25 = 0.16

- math (2 of 3) -
**drwls**, Thursday, August 7, 2008 at 8:55am<<(a) x^2+2x-3<(or equal to)0 >>

First solve

x^2 +2x -3 =0

(x+3)(x-1) = 0

x = -3 or 1

Between those points, for example at x=0, the function is negative.

(f(0)= -3, for example.)

-3 <_ x <_ 1 is the answer

- math (#3 of 3) -
**drwls**, Thursday, August 7, 2008 at 10:22am<<(2x-1)/(3x-2)<(or equal to)1>>

When equal to 1,

2x-1 = 3x-2

x = 1

If you go on the positive side if 1, for example at x = 2,

(2x-1)/(3x-2) = 3/5 which is <1

Any x >_ 1 satisfies the inequality

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