(a) If \(2^{x + y} = 16\) and \(4^{x - y} = \frac{1}{32}\), find the value of x and y. (b) P, Q and R are related in such a way that \(P \propto \frac{Q^{2}...
Question 1 Report
(a) If \(2^{x + y} = 16\) and \(4^{x - y} = \frac{1}{32}\), find the value of x and y.
(b) P, Q and R are related in such a way that \(P \propto \frac{Q^{2}}{R}\). When P = 36, Q = 3 and R = 4. Calculate Q when P = 200 and R = 2.
Answer Details
(a) Express both equations to base 2.
\(2^{x+y}=16=2^{4}\Rightarrow x+y=4\)
Since \(4=2^{2}\) and \(\frac{1}{32}=2^{-5}\): \(4^{x-y}=2^{2(x-y)}=2^{-5}\Rightarrow 2(x-y)=-5\Rightarrow x-y=-2.5\)
Solve the pair:
\[x+y=4,\qquad x-y=-2.5\]Adding: \(2x=1.5\Rightarrow x=0.75\). Then \(y=4-0.75=3.25\).
\(x=0.75,\ y=3.25\)
(b) \(P\propto\dfrac{Q^{2}}{R}\Rightarrow P=\dfrac{kQ^{2}}{R}\).
Using \(P=36,\ Q=3,\ R=4\): \(36=\dfrac{k(3)^{2}}{4}=\dfrac{9k}{4}\Rightarrow k=16\).
When \(P=200,\ R=2\): \(200=\dfrac{16Q^{2}}{2}=8Q^{2}\Rightarrow Q^{2}=25\Rightarrow Q=5\).
\(Q=5\)
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