If x=8^(2/3)*32^-(5/2)x=82332(52) then find xx?

3 Answers
May 31, 2018

See a solution process below:

Explanation:

Assuming the problem is:

x = (8^(2/3) * 32^(-5/2))x=(8233252)

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)xa×b=(xa)b

x = (8^(2 xx 1/3) * 32^(-5 xx 1/2))x=(82×13325×12)

x = ((8^2)^(1/3) * (32^-5)^(1/2))x=((82)13(325)12)

Next, use this rule of exponents to rewrite the 3232 term:

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

x = ((8^2)^(1/3) * ((1/32^(- -5))^(1/2))x=((82)13((1325)12)

x = ((8^2)^(1/3) * ((1/32^5)^(1/2))x=((82)13((1325)12)

x = ((64)^(1/3) * ((1/33554432)^(1/2))x=((64)13((133554432)12)

We can then use this rule to rewrite the exponents:

x^(1/color(red)(n)) = root(color(red)(n))(x)x1n=nx

x = root(3)(64) * 1/root(2)(33554432)x=3641233554432

x = root(3)(64) * 1/sqrt(33554432)x=364133554432

x = root(3)(64) * 1/sqrt(16777216 * 2)x=3641167772162

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))ab=ab

x = root(3)(64) * 1/(sqrt(16777216 * 2)x=3641167772162

x = 4 * 1/(sqrt(16777216)sqrt(2))x=41167772162

x = 4 * 1/(4096sqrt(2))x=4140962

x = 4/(4096sqrt(2))x=440962

x = 1/(1024sqrt(2))x=110242

Next, we can rationalize the fraction:

x = sqrt(2)/sqrt(2) * 1/(1024sqrt(2))x=22110242

x = (sqrt(2) * 1)/(sqrt(2) * 1024sqrt(2))x=21210242

x = sqrt(2)/(1024(sqrt(2))^2)x=21024(2)2

x = sqrt(2)/(1024 * 2)x=210242

x = sqrt(2)/(2048)x=22048

Or, approximately:

x = 0.00069x=0.00069

May 31, 2018

x=(8^(2/3)*32^(-5/2))x=(8233252)

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

x= ((root(3)8)^2*32^(-5/2))x=((38)23252)
x= (2^2*32^(-5/2))x=(223252)
x=(4*32^(-5/2))x=(43252)

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

x=(4*1/(32^(5/2)))x=(413252)

x=(4*1/(sqrt32)^5)x=⎜ ⎜41(32)5⎟ ⎟

x=(4*1/(sqrt(16xx2))^5)x=⎜ ⎜41(16×2)5⎟ ⎟

x=(4*1/(4sqrt2)^5)x=⎜ ⎜41(42)5⎟ ⎟

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

x=(4*1/(4^5*(sqrt2)^5))x=⎜ ⎜4145(2)5⎟ ⎟

x=(4*1/(4^5*(2^((1/2)*5))))x=⎜ ⎜ ⎜4145(2(12)5)⎟ ⎟ ⎟

x=(4*1/(4^5*(2^(5/2))))x=4145(252)

x=(4*1/((2^2)^5*(2^(5/2))))x=41(22)5(252)

2^(5/2)= 2^(2+1/2)= 2^2*2^(1/2)= 2^2*sqrt2252=22+12=22212=222

(2^2)^5=(2^10)(22)5=(210)

x=(4*1/((2^10*2^2*sqrt2)))x=41(210222)

x=(4/(2^12*sqrt2))x=(42122)

x=(2^2/(2^12*sqrt2))x=(222122)

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