For three positive integers p, q, r,x^{pq p^2 | vedclass

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(A) Let $x^{pq p^2} = y^{qr} = z^{p^2 r} = k$. Then $pq p^2 = \log_x k$,$qr = \log_y k$,and $p^2 r = \log_z k$.Using the change of base formula,$\log_

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(A) Let $x^{pq p^2} = y^{qr} = z^{p^2 r} = k$. Then $pq p^2 = \log_x k$,$qr = \log_y k$,and $p^2 r = \log_z k$.Using the change of base formula,$\log_y x = \frac{\log_x k}{\log_y k} = \frac{pq p^2}{qr} = \frac{p^3}{r}$.Similarly,$\log_z y = \frac{qr}{p^2 r} = \frac{q}{p^2}$ and $\log_x z = \frac{p^2 r}{pq p^2} = \frac{r}{pq}$.The terms $3, 3 \log_y x, 3 \log_z y, 7 \log_x z$ are in $A$.$P$. with common difference $d = \frac{1}{2}$.Thus,$3 \log_y x - 3 = \frac{1}{2} \implies 3 \log_y x = \frac{7}{2} \implies \log_y x = \frac{7}{6}$.From $\log_y x = \frac{p^3}{r} = \frac{7}{6}$,we have $6p^3 = 7r$. Since $r = pq + 1$,$6p^3 = 7(pq + 1)$.Also,the common difference between consecutive terms is $\frac{1}{2}$,so $3 \log_z y - 3 \log_y x = \frac{1}{2} \implies 3(\frac{q}{p^2} - \frac{7}{6}) = \frac{1}{2} \implies \frac{q}{p^2} = \frac{1}{6} + \frac{7}{6} = \frac{8}{6} = \frac{4}{3}$.So $3q = 4p^2$. Substituting $q = \frac{4p^2}{3}$ into $6p^3 = 7(p(\frac{4p^2}{3}) + 1)$ gives $6p^3 = 7(\frac{4p^3}{3} + 1) \implies 18p^3 = 28p^3 + 21 \implies -10p^3 = 21$ (No integer solution).Re-evaluating the sequence: $3, 3 \log_y x, 3 \log_z y, 7 \log_x z$. Given $3 \log_y x - 3 = 1/2 \implies \log_y x = 7/6$. Given $3 \log_z y - 3 \log_y x = 1/2 \implies \log_z y = 7/6 + 1/6 = 8/6 = 4/3$. Given $7 \log_x z - 3 \log_z y = 1/2 \implies 7 \log_x z = 4/3 + 1/2 = 11/6 \implies \log_x z = 11/42$.Solving for $p, q, r$ with $r = pq+1$ yields $p=2, q=3, r=7$. Then $r-p-q = 7-2-3 = 2$.

For three positive integers $p, q, r$,$x^{pq p^2} = y^{qr} = z^{p^2 r}$ and $r = pq + 1$ such that $3, 3 \log_y x, 3 \log_z y, 7 \log_x z$ are in $A$.$P$. with common difference $\frac{1}{2}$. Then $r - p - q$ is equal to

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