What is the value of#x# if#x^(4/5)=(2^8)/(3^8)#?

3 Answers
Mar 18, 2017

#x=1024/59049#

Explanation:

If #x > 0# and #a, b# are any real numbers, then:

#(x^a)^b = x^(ab)#

So we find:

#x = x^1 = x^(4/5*5/4) = (x^(4/5))^(5/4)=(2^8/3^8)^(5/4)= ((2/3)^8)^(5/4) = (2/3)^(8*5/4) = (2/3)^10 = 2^10/3^10=1024/59049#

Mar 18, 2017

#x=(2/3)^10#

Explanation:

In general if #x^a=p^b# then
#color(white)("XXX")x=(x^a)^(1/a) = (p^b)^(1/a)#

In this case
#color(white)("XXX")a = 4/5color(white)("XX")rarrcolor(white)("XX")1/a=5/4#

#color(white)("XXX")p=2/3#
and
#color(white)("XXXXXX")#after noting that #(2^8)/(3^8)=(2/3)^8#
#color(white)("XXX")b=8#

So
#color(white)("XXX")x=((2/3)^8)^(5/4) = (2/3)^((8 * 5)/4) = (2/3)^10#

Mar 18, 2017

#x=0.01734152#

Explanation:

#x^(4/5)=(2^8)/3^8#

Make #x# radical,

#root(5)(x^4)=2^8/3^8#

Multiply both sides by the index of 5,

#(root(5)(x^4))^5=(2^8/3^8)^5#
#x^4=2^40/3^40#

Root both sides by index of 4,

#root(4)(x^4)=root(4)(2^40/3^40)#
#(x^4)^(1/4)=(2^40/3^40)^(1/4)#
#x=2^10/3^10#
#x=1024/59049#

Hence #x=0.01734152#.