(a) Make q the subject of the relation \(t = \sqrt{\frac{pq}{r} - r^{2}q}\). (b) If \(9^{(1 - x)} = 27^{y}\) and \(x - y = -1\frac{1}{2}\), find the value o...
Question 1 Report
(a) Make q the subject of the relation \(t = \sqrt{\frac{pq}{r} - r^{2}q}\).
(b) If \(9^{(1 - x)} = 27^{y}\) and \(x - y = -1\frac{1}{2}\), find the value of x and y.
Answer Details
(a) Square both sides to remove the root:
\[t^2 = \frac{pq}{r} - r^2 q = q\left(\frac{p}{r} - r^2\right) = q\left(\frac{p - r^3}{r}\right).\]
Make q the subject:
\[q = \frac{r\,t^2}{p - r^3}.\]
(b) Write both sides to base 3: \(9^{(1-x)} = 3^{2(1-x)}\) and \(27^{y} = 3^{3y}\). Equating indices,
\[2(1 - x) = 3y \Rightarrow 2 - 2x = 3y. \quad (1)\]
Also \(x - y = -\tfrac{3}{2}\Rightarrow x = y - \tfrac{3}{2}. \quad (2)\)
Substitute (2) into (1):
\[2 - 2\left(y - \tfrac{3}{2}\right) = 3y \Rightarrow 2 - 2y + 3 = 3y \Rightarrow 5 = 5y \Rightarrow y = 1.\]
\[x = 1 - \tfrac{3}{2} = -\tfrac{1}{2}.\]
Answer: \(x = -\tfrac{1}{2},\ y = 1\).
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