Bunuel wrote:
Let \(p = 3^{(q+1)}\) and \(q = 2r\). Then \(\frac{p}{3^2} =\)
(A) 3^(2r−1)
(B) 3^(2r)
(C) 3
(D) r
(E) 3^(2r+1)
\(p = 3^{(q+1)}\)
Given;\(q = 2r\)
Substituting value of q in\(p = 3^{(q+1)}\) weget;
\(p = 3^{(2r+1)}\)
Dividing both sides by\(3^2\)
\(\frac{p}{3^2}\) =\(\frac{3^{(2r+1)}}{3^2}\)
\(\frac{p}{3^2}\) =\(3^{(2r+1)} * 3^{(-2)}\)
\(\frac{p}{3^2}\) =\(3^{(2r+1-2)}\)
\(\frac{p}{3^2}\) =\(3^{(2r-1)}\)
Answer A...
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