Question 1061717: Without using a calculator, order the following numbers from least to greatest: A=(2^1/2)/(4^1/6). B=12th root of 128. C=(1/(8^ 1/5))^2. D=square root of (4^-1)/(2^-1)*(8^-1). E=cube root of (2^1/2)*(4^-1/4)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Make them all powers of 2
The first is 2^(1/2)*(2^2)^1/6=2^5/6
The second is 2^(7/12), because 2^7=128
The third is 8^(-1/5)^2=8^(-2/5)=(2^3)^(-2/5)=2^*(-6/5)
The fourth is sqrt ((1/4)/(1/2)(1/8)=sqrt (1/4)/(1/16)=sqrt (4)=2.
The fifth is 2^(1/2)(2^(-1/2)), because 4^(-1/4) is 1/4^(1/4)=1/(2^2)*(1/4)=1/2^(2/4)=1/2^(1/2)=2^(-1/2)=2^0=1
C is the smallest. 2^(-5/6) is less than 1
then E, for 2^0=1
then B, because 2^(7/12) is greater than square root of 2.
then A, because 2^(5/6) is almost 2 and is 2^(10/12)
then D, 2
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