X4+x2+25

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$\\Solution:\\ Given:\\\\ 4^x-4^{x-1}=24\\\\ \Rightarrow 4^x-\frac{4^x}{4}=24\ \ \ \ [\because a^{m-n}=\frac{a^m}{a^n}]\\\\\\ \Rightarrow 4^x(1-\frac{1}4)=24\\\\\ \Rightarrow 4^x\times\frac34=24\\\\\\ \Rightarrow 4^x=24\times\frac43\\\\\\\ \Rightarrow 4^x=32\\\\\\ \Rightarrow {(2^2)}^x=2^5\\\\ Bases\ are\ equal\ then\ the\ exponents\ are\ also\ equal\\\\ \Rightarrow 2x=5\\\\ \therefore x=\frac52\\\\ Now, \\ (2x)^x={(2\times\frac52)}^\frac52 =5^\frac52=\sqrt{5^5}=\sqrt{5^4\times5}=5^2\sqrt5=25\sqrt5$

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$\\Solution:\\\\ Given: x+\frac1x=5\\\\ \text{Squaring on both sides}\\\\ \Rightarrow (x+\frac1x)^2=5^2\\\\ \Rightarrow x^2+2\times x\times\frac1x+\left (\frac{1}{x} \right )^2=25\ \ \ \ \ \ \ \ [\because (a+b)^2=a^2+b^2+2ab]\\\\ \Rightarrow x^2+2+\frac{1}{x^2}=25\\\\ \Rightarrow x^2+\frac{1}{x^2}=25-2\\\\ \therefore \left (x^2+\frac{1}{x^2} \right )=23\\\\ \text{Squaring on both sides}\\\\ \Rightarrow \left (x^2+\frac{1}{x^2} \right )^2=23^2\\ \Rightarrow x^4+2\times x^2\times\frac1{x^2}+\left (\frac{1}{x^2} \right )^2=529\ \ \ \ \ \ \ \ [\because (a+b)^2=a^2+b^2+2ab]\\\\ \Rightarrow x^4+2+\frac{1}{x^4}=529\\\\ \Rightarrow x^4+\frac{1}{x^4}=529-2\\\\ \therefore \left (x^4+\frac{1}{x^4} \right )=527$

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