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Question

Question: If \[x\] varies as the \[{m^{th}}\] power of \[y\], \[y\] varies as the \[{n^{th}}\] power of \[z\] ...

If xx varies as the mth{m^{th}} power of yy, yy varies as the nth{n^{th}} power of zz and xx varies as the pth{p^{th}} power of zz, then which one of the following is correct?
A. p=m+np = m + n
B. p=mnp = m - n
C. p=mnp = mn
D. None of the above

Explanation

Solution

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,np,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 xx varies as the mth{m^{th}} power of yy i.e., y=x1mx=ymy = {x^{\dfrac{1}{m}}} \Rightarrow x = {y^m}
And yy varies as the nth{n^{th}} power of zz i.e., z=y1ny=znz = {y^{\dfrac{1}{n}}} \Rightarrow y = {z^n}

y=zn x=(ym)n [x=ym] x=zmn.................................(1)  \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 xx varies as the pth{p^{th}} power of zzi.e., z=x1px=zp.........................(2)z = {x^{\dfrac{1}{p}}} \Rightarrow x = {z^p}.........................\left( 2 \right)
From equation (1) and (2), we have
xmn=xp\Rightarrow {x^{mn}} = {x^p}
Since, the bases are equal we can equate the powers on both sides

mn=p p=mn  \Rightarrow mn = p \\\ \therefore p = mn \\\

Thus, the correct option is C. p=mnp = mn

Note : Here, if aa varies as the bth{b^{th}} power of cc, then it can be written as a=c1bc=aba = {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 xm=xn{x^m} = {x^n} then m=nm = n.