Solve the following equation for x:
(43)2x+12=132(\sqrt[3]{4})^{2x + \dfrac{1}{2}} = \dfrac{1}{32}(34)2x+21=321
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Given,
⇒(43)2x+12=132⇒[(22)13]2x+12=125⇒(22)2x3+16=2−5⇒(2)4x3+13=2−5⇒4x+13=−5⇒4x+1=−15⇒4x=−16⇒x=−4.\Rightarrow (\sqrt[3]{4})^{2x + \dfrac{1}{2}} = \dfrac{1}{32} \\[1em] \Rightarrow [(2^2)^{\dfrac{1}{3}}]^{2x + \dfrac{1}{2}} = \dfrac{1}{2^5} \\[1em] \Rightarrow (2^2)^{\dfrac{2x}{3} + \dfrac{1}{6}} = 2^{-5} \\[1em] \Rightarrow (2)^{\dfrac{4x}{3} + \dfrac{1}{3}} = 2^{-5} \\[1em] \Rightarrow \dfrac{4x + 1}{3} = -5 \\[1em] \Rightarrow 4x + 1 = -15 \\[1em] \Rightarrow 4x = -16 \\[1em] \Rightarrow x = -4.⇒(34)2x+21=321⇒[(22)31]2x+21=251⇒(22)32x+61=2−5⇒(2)34x+31=2−5⇒34x+1=−5⇒4x+1=−15⇒4x=−16⇒x=−4.
Hence, x = -4.
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(13)x=44−34−6(13)^{\sqrt{x}} = 4^4 - 3^4 - 6(13)x=44−34−6
(35)x+1=12527\Big(\sqrt{\dfrac{3}{5}}\Big)^{x + 1} = \dfrac{125}{27}(53)x+1=27125
pq=(qp)1−2x\sqrt{\dfrac{p}{q}} = \Big(\dfrac{q}{p}\Big)^{1 - 2x}qp=(pq)1−2x
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
