Questions & Answers

Question

Answers

A. \[p = m + n\]

B. \[p = m - n\]

C. \[p = mn\]

D. None of the above

Answer
Verified

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 \\

\]