# 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 ->  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

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 Click here to see ALL problems on Exponents 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(7569)   (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(57377)   (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.