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1.-x^(2)y^(2)/x^(1/2)y^(0)

Writes in expinential form and simplify.

^3sqrt x^(5)y^(5)

^3sqrt x^(2)y

^3sqrt a*^5sqrt a

3. Solve the equation.
2x^(-1/2)=8

4. If f(x) =3x+1, and g(x)=(x^(2)+5x)^(-1/2), find g (f(x)).

5. log4, 32=x

6.log10, 0.01=-2

7. loga, A=1

2007-05-05 11:30:05 · 1 answers · asked by Willy J 1 in Science & Mathematics Mathematics

1 answers

1. It's not entirely clear what you're looking for. The equations simplified are (assuming "^3sqrt" is "cube root")...

- x^(3/2)y^2
x^(5/3)y^(5/3)
x^(2/3)y^(1/3)
a^(13/6)

3. 2x^(-1/2) = 8
x^(-1/2) = 4
1/sqrt(x) = 4
sqrt(x) = 1/4
x = 1/16

4. g(f(x)) = ((3x+1)^2 + 5(3x+1))^(-1/2)
= (9x^2 + 6x + 1 + 15x + 5)^(-1/2)
= (9x^2 + 21x + 6)^(-1/2)
= 1/sqrt(9x^2 + 21x + 6)

5. log4(32) = x
4^x = 32
2^(2x) = 32
2x = 5
x = 2.5

6. log10(.01) = -2
Not sure what you want done here, that's simply a true statement, nothing to solve.

7. loga(A) = 1
A = a^1
A = a

2007-05-05 11:45:53 · answer #1 · answered by McFate 7 · 0 0

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