Solve the following equation for x:
(35)x+1=12527\Big(\sqrt{\dfrac{3}{5}}\Big)^{x + 1} = \dfrac{125}{27}(53)x+1=27125
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Given,
⇒(35)x+1=12527⇒(35)x+12=(53)3⇒(35)x+12=(35)−3∴x+12=−3⇒x+1=−6⇒x=−7.\Rightarrow \Big(\sqrt{\dfrac{3}{5}}\Big)^{x + 1} = \dfrac{125}{27} \\[1em] \Rightarrow \Big(\dfrac{3}{5}\Big)^{\dfrac{x + 1}{2}} = \Big(\dfrac{5}{3}\Big)^3 \\[1em] \Rightarrow \Big(\dfrac{3}{5}\Big)^{\dfrac{x + 1}{2}} = \Big(\dfrac{3}{5}\Big)^{-3} \\[1em] \therefore \dfrac{x + 1}{2} = -3 \\[1em] \Rightarrow x + 1 = -6 \\[1em] \Rightarrow x = -7.⇒(53)x+1=27125⇒(53)2x+1=(35)3⇒(53)2x+1=(53)−3∴2x+1=−3⇒x+1=−6⇒x=−7.
Hence, x = -7.
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52x + 3 = 1
(13)x=44−34−6(13)^{\sqrt{x}} = 4^4 - 3^4 - 6(13)x=44−34−6
(43)2x+12=132(\sqrt[3]{4})^{2x + \dfrac{1}{2}} = \dfrac{1}{32}(34)2x+21=321
pq=(qp)1−2x\sqrt{\dfrac{p}{q}} = \Big(\dfrac{q}{p}\Big)^{1 - 2x}qp=(pq)1−2x
