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

3 Answers
smendyka
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#

Layomi.O
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#

EZ as pi
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#

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