For three positive integers p, q, r, x^pq^2 = y^qr = z^p^2r and r = pq + 1 such that 3, 3 …

Question

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

Accepted Answer

$x^{p q^{2}}=y^{q r}=z^{p^{2} r}$

$$
3 \log _{y} x=\frac{7}{2}, 3 \log _{z} y=4,7 \log _{x} z=\frac{9}{2}
$$

$$
\begin{aligned}
& \Rightarrow x=y^{\frac{7}{6}}, y=z^{\frac{4}{3}}, z=x^{\frac{9}{14}} \\
& y^{\frac{7}{6} p q^{2}}=y^{q r}=y^{\frac{3}{4} p^{2} r} \\
& \Rightarrow \frac{7}{6} p q^{2}=q r=\frac{3}{4} p^{2} r \\
& \therefore 7 p q=6 r, 4 q=3 p^{2} \\
& r=p q+1 \\
& r=\frac{6 r}{7}+1 \Rightarrow r=7 \\
& p q=6 \\
& p\left(\frac{3 p^{2}}{4}\right)=6 \\
& p=2, q=3 \\
& r-p-q=7-5=2
\end{aligned}
$$

Solutions · 1