log 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
Solve for x (without using a calculator):
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!
3 answers
<<(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
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
<<(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
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
3 answers