Question

If x and y are whole numbers such that
${y}^{x}$
= 19683 and y > x and 1 < x < 4, then
$\sqrt[x]{y}$
is __

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Answer 1

Answer:

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Answer 2

Answer:

19683 is 3^8

x is either 2 or 3 from the condition given in the question.

Therefore, square of any integer does not end on 2.

So,cube is the only possibility.

It implies 27^3 =19683.

It follows cube root of 27 is 3 where x is 3 and y is 27.