Given: root(3)(k^d)=(root(3)(k))^53√kd=(3√k)5
Cube both sides
k^d=(root(3)(k))^15kd=(3√k)15
color(brown)("Doing it the 'hard way'")Doing it the 'hard way'
Write (root(3)(k))^15(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)][3√k×3√k×3√k]×[3√k×3√k×3√k]×[3√k×3√k×3√k]×[3√k×3√k×3√k]×[3√k×3√k×3√k]
Which is the same as k^5 k5
Putting it all back together we have:
k^d=k^5kd=k5
Then by direct comparison we have:
d=5d=5
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color(brown)("Doing it the more straight forward way")Doing it the more straight forward way
Write (root(3)(k))^15(3√k)15 as k^((15/3)k(153)
but 15-:3 = 515÷3=5 giving
(root(3)(k))^15(3√k)15 is the same as k^5k5
Then the rest is as above.
