If x=8^(2/3)*32^-(5/2) then find x?

3 Answers
May 31, 2018

See a solution process below:

Explanation:

Assuming the problem is:

x = (8^(2/3) * 32^(-5/2))

First, use this rule of exponents to rewrite each term on the right side:

x^(color(red)(a) xx color(blue)(b)) = (x^color(red)(a))^color(blue)(b)

x = (8^(2 xx 1/3) * 32^(-5 xx 1/2))

x = ((8^2)^(1/3) * (32^-5)^(1/2))

Next, use this rule of exponents to rewrite the 32 term:

x^color(red)(a) = 1/x^color(red)(-a)

x = ((8^2)^(1/3) * ((1/32^(- -5))^(1/2))

x = ((8^2)^(1/3) * ((1/32^5)^(1/2))

x = ((64)^(1/3) * ((1/33554432)^(1/2))

We can then use this rule to rewrite the exponents:

x^(1/color(red)(n)) = root(color(red)(n))(x)

x = root(3)(64) * 1/root(2)(33554432)

x = root(3)(64) * 1/sqrt(33554432)

x = root(3)(64) * 1/sqrt(16777216 * 2)

Now, use this rule to rewrite the denominator of the fraction:

sqrt(color(red)(a) * color(blue)(b)) = sqrt(color(red)(a)) * sqrt(color(blue)(b))

x = root(3)(64) * 1/(sqrt(16777216 * 2)

x = 4 * 1/(sqrt(16777216)sqrt(2))

x = 4 * 1/(4096sqrt(2))

x = 4/(4096sqrt(2))

x = 1/(1024sqrt(2))

Next, we can rationalize the fraction:

x = sqrt(2)/sqrt(2) * 1/(1024sqrt(2))

x = (sqrt(2) * 1)/(sqrt(2) * 1024sqrt(2))

x = sqrt(2)/(1024(sqrt(2))^2)

x = sqrt(2)/(1024 * 2)

x = sqrt(2)/(2048)

Or, approximately:

x = 0.00069

May 31, 2018

x=(8^(2/3)*32^(-5/2))

Using the law of indices: a^(m/n)=(root(n)a)^m

x= ((root(3)8)^2*32^(-5/2))
x= (2^2*32^(-5/2))
x=(4*32^(-5/2))

Using the law of indices a^-n=1/(a^n)

x=(4*1/(32^(5/2)))

x=(4*1/(sqrt32)^5)

x=(4*1/(sqrt(16xx2))^5)

x=(4*1/(4sqrt2)^5)

Using the law of indices: (ab)^m=a^m*b^m

x=(4*1/(4^5*(sqrt2)^5))

x=(4*1/(4^5*(2^((1/2)*5))))

x=(4*1/(4^5*(2^(5/2))))

x=(4*1/((2^2)^5*(2^(5/2))))

2^(5/2)= 2^(2+1/2)= 2^2*2^(1/2)= 2^2*sqrt2

(2^2)^5=(2^10)

x=(4*1/((2^10*2^2*sqrt2)))

x=(4/(2^12*sqrt2))

x=(2^2/(2^12*sqrt2))

x=(cancel(2^2)/(cancel(2^12)*(sqrt2)))

#x=(1/(2^10*sqrt2))

x=(1/(1024sqrt2))

Rationalise

x=1/(1024sqrt2)*sqrt2/sqrt2

x=sqrt2/((1024sqrt2)sqrt2)

x=sqrt2/(1024xx2)

therefore x=sqrt2/2048

May 31, 2018

x = sqrt2/2048

Explanation:

x = 8^(2/3) xx 32^(-5/2)

Both 8 and 32 are powers of 2

x = (2^3)^(2/3) xx (2^5)^(-5/2) ' "larr multiply the indices

x =2^2 xx 2^(-25/2)

x = 2^(2-25/2)

x = 2^(-21/2)

x = 1/(2^(21/2))

x = 1/(sqrt2)^21

Rationalise the denominator

x =1/(sqrt2)^21 xx sqrt2/sqrt2

x = sqrt2/sqrt2^22

x =sqrt2/2^11

x = sqrt2/2048

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

However, consider if the question had been:

x = 8^(2/3) xx 32^(-2/5)

x = (2^3)^(2/3) xx (2^5)^(-2/5)

x = 2^2 xx 2^-2

x = 2^0

x =1