Question 551638: solve for x
(8x)^-3=64
(4-x)^(1/2)=8
Simplify
(2r^-1)^4 * (4r^2)^-2
(4x^-3)^(-1/2) * 4x^(1/2)
(8a^-6)^(-2/3) Found 2 solutions by lwsshak3, Maths68:Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! solve for x
..
(8x)^-3=64
1/(8x)^3=64
take cube root of both sides
1/8x=4
32x=1
x=1/32
..
(4-x)^(1/2)=8
√(4-x)=8
square both sides
4-x=64
x=-60
..
Simplify
(2r^-1)^4 * (4r^2)^-2
2^4*r^-4*4^-2*r^-4
2^4r^-4*2^-4r^-4
2^0*r^-8=r^-8
..
(4x^-3)^(-1/2) * 4x^(1/2)
1/√(4x^-3) *√4x
√4x/√(4x^-3)
√[4x/(4x^-3)]
√(x/x^-3)
√(x^4)=x^2
..
(8a^-6)^(-2/3)
8^-2/3*a^4
1/8^2/3*a^4
1/4*a^4
a^4/4
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