What is the value of #d# in the equation #root3 (k^d) = (root3 k)^5#?

2 Answers

Given: #root(3)(k^d)=(root(3)(k))^5#

Cube both sides

#k^d=(root(3)(k))^15#

#color(brown)("Doing it the 'hard way'")#

Write #(root(3)(k))^15# as

#[ root(3)(k) xxroot(3)(k)xxroot(3)(k)] xx[ root(3)(k) xxroot(3)(k)xxroot(3)(k)] xx[ root(3)(k) xxroot(3)(k)xxroot(3)(k)] xx[ root(3)(k) xxroot(3)(k)xxroot(3)(k)]xx[ root(3)(k) xxroot(3)(k)xxroot(3)(k)]#

Which is the same as #k^5 #

Putting it all back together we have:

#k^d=k^5#

Then by direct comparison we have:

#d=5#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(brown)("Doing it the more straight forward way")#

Write #(root(3)(k))^15# as #k^((15/3)#

but #15-:3 = 5# giving

#(root(3)(k))^15# is the same as #k^5#

Then the rest is as above.

Aug 7, 2018

#color(orange)(d = 5#

Explanation:

#root3 k^d = (root 3 k)^5#

#k^(d/3) = (k^(1/3))^5#

#k^(d/3) = k^(5/3)#

#d /cancel 3 = 5/cancel3#

#color(orange)(d = 5#