Question

# If $x$ varies as the ${m^{th}}$ power of $y$, $y$ varies as the ${n^{th}}$ power of $z$ and $x$ varies as the ${p^{th}}$ power of $z$, then which one of the following is correct?A. $p = m + n$B. $p = m - n$C. $p = mn$D. None of the above

Hint: In this question, we will proceed by writing the given data and converting them to the desired way. Then substitute the terms in each other to form a relation between $p,m,n$ to get the required answer. So, use this concept to reach the solution of the given problem.

Given that $x$ varies as the ${m^{th}}$ power of $y$ i.e., $y = {x^{\dfrac{1}{m}}} \Rightarrow x = {y^m}$
And $y$ varies as the ${n^{th}}$ power of $z$ i.e., $z = {y^{\dfrac{1}{n}}} \Rightarrow y = {z^n}$
$\Rightarrow y = {z^n} \\ \Rightarrow x = {\left( {{y^m}} \right)^n}{\text{ }}\left[ {\because x = {y^m}} \right] \\ \Rightarrow x = {z^{mn}}.................................\left( 1 \right) \\$
Also given that and $x$ varies as the ${p^{th}}$ power of $z$i.e., $z = {x^{\dfrac{1}{p}}} \Rightarrow x = {z^p}.........................\left( 2 \right)$
$\Rightarrow {x^{mn}} = {x^p}$
$\Rightarrow mn = p \\ \therefore p = mn \\$
Thus, the correct option is C. $p = mn$
Note: Here, if $a$ varies as the ${b^{th}}$ power of $c$, then it can be written as $a = {c^{\dfrac{1}{b}}} \Rightarrow c = {a^b}$. Whenever we have equal bases on both sides, we can equate the powers of the terms on both sides i.e., if ${x^m} = {x^n}$ then $m = n$.