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

Answer 1

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.

Complete step-by-step answer:
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., \[z = {x^{\dfrac{1}{p}}} \Rightarrow x = {z^p}.........................\left( 2 \right)\]
From equation (1) and (2), we have
\[ \Rightarrow {x^{mn}} = {x^p}\]
Since, the bases are equal we can equate the powers on both sides
\[
   \Rightarrow mn = p \\
  \therefore p = mn \\
\]
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 \[m = n\].