Solve the following equation for x:
pq=(qp)1−2x\sqrt{\dfrac{p}{q}} = \Big(\dfrac{q}{p}\Big)^{1 - 2x}qp=(pq)1−2x
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Given,
⇒pq=(qp)1−2x⇒(pq)12=(pq)−(1−2x)⇒(pq)12=(pq)(2x−1)⇒12=2x−1⇒2x=1+12⇒2x=32⇒x=34.\Rightarrow \sqrt{\dfrac{p}{q}} = \Big(\dfrac{q}{p}\Big)^{1 - 2x} \\[1em] \Rightarrow \Big(\dfrac{p}{q}\Big)^{\dfrac{1}{2}} = \Big(\dfrac{p}{q}\Big)^{-(1 - 2x)} \\[1em] \Rightarrow \Big(\dfrac{p}{q}\Big)^{\dfrac{1}{2}} = \Big(\dfrac{p}{q}\Big)^{(2x - 1)} \\[1em] \Rightarrow \dfrac{1}{2} = 2x - 1 \\[1em] \Rightarrow 2x = 1 + \dfrac{1}{2} \\[1em] \Rightarrow 2x = \dfrac{3}{2} \\[1em] \Rightarrow x = \dfrac{3}{4}.⇒qp=(pq)1−2x⇒(qp)21=(qp)−(1−2x)⇒(qp)21=(qp)(2x−1)⇒21=2x−1⇒2x=1+21⇒2x=23⇒x=43.
Hence, x = 34\dfrac{3}{4}43.
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(35)x+1=12527\Big(\sqrt{\dfrac{3}{5}}\Big)^{x + 1} = \dfrac{125}{27}(53)x+1=27125
(43)2x+12=132(\sqrt[3]{4})^{2x + \dfrac{1}{2}} = \dfrac{1}{32}(34)2x+21=321
4x−1×(0.5)3−2x=(18)x4^{x - 1} \times (0.5)^{3 - 2x} = \Big(\dfrac{1}{8}\Big)^x4x−1×(0.5)3−2x=(81)x
If 53x = 125 and (10)y = 0.001, find x and y.
