SOLUTION: Please help :) 1.Evaluate: a) (8/27)^(-4/3) b)[(3/5)^-1 - 4^0] ^-2 -- I got 9/4. 2.Which is equivalent to: a)[(25^x)]/[125^(x+1)(5)] Is it 5^x+2 ? b) [(9^x)]/[(27^2x)(3

Algebra ->  Exponents -> SOLUTION: Please help :) 1.Evaluate: a) (8/27)^(-4/3) b)[(3/5)^-1 - 4^0] ^-2 -- I got 9/4. 2.Which is equivalent to: a)[(25^x)]/[125^(x+1)(5)] Is it 5^x+2 ? b) [(9^x)]/[(27^2x)(3      Log On


   

Question 271578: Please help :)
1.Evaluate:
a) (8/27)^(-4/3)
b)[(3/5)^-1 - 4^0] ^-2 -- I got 9/4.
2.Which is equivalent to:
a)[(25^x)]/[125^(x+1)(5)] Is it 5^x+2 ?
b) [(9^x)]/[(27^2x)(3)] Is it 3^4x-1 ?
c) (1/36)^x-2 Is it 6^-2x+4 ?
Found 2 solutions by edjones, stanbon:
Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
b)[(3/5)^-1 - 4^0] ^-2
=[5/3 - 1]^-2
=[2/3]^-2
=[3/2]^2
=9/4 Very good.
.
Ed

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
1.Evaluate:
a) (8/27)^(-4/3)
Take the cube root of (8/27) to get:
= (2/3)^-4
= (3/2)^4
= 81/16
---------------------------------------
b)[(3/5)^-1 - 4^0] ^-2 -- I got 9/4.
= [(5/3) - 1]^-2
= [2/3]^-2
= [3/2)^2
= 9/4
You Got It.
-------------------------------------



2.Which is equivalent to:
a)[(25^x)]/[125^(x+1)(5)] Is it 5^x+2 ?
= [5^2x]/[5^(3x+3)*5]
= [5^2x]/[5^(3x+4)]
= 5^(-x-4)
-------------------------------------


b) [(9^x)]/[(27^2x)(3)] Is it 3^4x-1 ?
= [3^2x]/[3^6x*3]
= [3^2x]/[3^(6x+1)]
= 3^(-4x-1)
--------------------------------------

c) (1/36)^(x-2) Is it 6^-2x+4 ?
= (6^-2)^(x-2)
= 6^(-2x+4)
You Got It but be careful; use parentheses to avoid confusion.
===================
Cheers,
Stan H.