Solve the following equation for x:
(13)x=44−34−6(13)^{\sqrt{x}} = 4^4 - 3^4 - 6(13)x=44−34−6
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Given,
⇒(13)x=44−34−6⇒(13)x=256−81−6⇒(13)x=169⇒(13)x=132∴x=2⇒x=4.\Rightarrow (13)^{\sqrt{x}} = 4^4 - 3^4 - 6 \\[1em] \Rightarrow (13)^{\sqrt{x}} = 256 - 81 - 6 \\[1em] \Rightarrow (13)^{\sqrt{x}} = 169 \\[1em] \Rightarrow (13)^{\sqrt{x}} = 13^2 \\[1em] \therefore \sqrt{x} = 2 \\[1em] \Rightarrow x = 4.⇒(13)x=44−34−6⇒(13)x=256−81−6⇒(13)x=169⇒(13)x=132∴x=2⇒x=4.
Hence, x = 4.
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If (p−1q2p2q−4)7÷(p3q−5p−2q3)−5=pxqy\Big(\dfrac{p^{-1}q^{2}}{p^2q^{-4}}\Big)^7 ÷ \Big(\dfrac{p^3q^{-5}}{p^{-2}q^{3}}\Big)^{-5} = p^xq^y(p2q−4p−1q2)7÷(p−2q3p3q−5)−5=pxqy, find x + y, where p and q are different positive primes.
52x + 3 = 1
(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
