Question

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

Answer 1

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`

Answer 2

`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`

Answer 3

`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`