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Question Number 82665 by jagoll last updated on 23/Feb/20

(√(6561(√(6561(√(6561(√(6561(√(...)))))))))) = 3^( 8^x ) (√(6561(√(...)))) (60 time) find x

$$\sqrt{\mathrm{6561}\sqrt{\mathrm{6561}\sqrt{\mathrm{6561}\sqrt{\mathrm{6561}\sqrt{...}}}}}\:=\:\mathrm{3}^{\:\mathrm{8}^{{x}} \:} \\ $$$$\sqrt{\mathrm{6561}\sqrt{...}}\:\:\left(\mathrm{60}\:{time}\right) \\ $$$${find}\:{x} \\ $$

Commented by mr W last updated on 23/Feb/20

6561^((1/2)+(1/2^2 )+(1/2^3 )+...+(1/2^(60) )) =3^8^x 6561^((1/2)(((1−(1/2^(60) ))/(1−(1/2))))) =3^8^x 6561^(1−(1/2^(60) )) =3^8^x 3^(8(1−(1/2^(60) ))) =3^8^x 8(1−(1/2^(60) ))=8^x 1−(1/2^(60) )=8^(x−1) =2^(3(x−1)) ⇒x=1+(1/3)log_2 (1−(1/2^(60) ))

$$\mathrm{6561}^{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+...+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{60}} }} =\mathrm{3}^{\mathrm{8}^{{x}} } \\ $$$$\mathrm{6561}^{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{60}} }}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}\right)} =\mathrm{3}^{\mathrm{8}^{{x}} } \\ $$$$\mathrm{6561}^{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{60}} }} =\mathrm{3}^{\mathrm{8}^{{x}} } \\ $$$$\mathrm{3}^{\mathrm{8}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{60}} }\right)} =\mathrm{3}^{\mathrm{8}^{{x}} } \\ $$$$\mathrm{8}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{60}} }\right)=\mathrm{8}^{{x}} \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{60}} }=\mathrm{8}^{{x}−\mathrm{1}} =\mathrm{2}^{\mathrm{3}\left({x}−\mathrm{1}\right)} \\ $$$$\Rightarrow{x}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{log}_{\mathrm{2}} \:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{60}} }\right) \\ $$

Commented by jagoll last updated on 23/Feb/20

thank you mister w and mister john

$${thank}\:{you}\:{mister}\:{w}\:{and}\:{mister}\:{john} \\ $$

Answered by john santu last updated on 23/Feb/20

3^((4/2)+(4/4)+(4/8)+(4/(16))+...+(4/2^(60) )) = 3^8^(x ) ⇒ 4 ((1/2)+(1/4)+(1/8)+...+(1/2^(60) )) = 8^x ⇒4(((((1/2))(1−((1/2))^(60) ))/(1−(1/2)))) = 8^x ⇒1−2^(−60) = 2^(3x−2) ⇒ 3x−2 = log_2 (1−2^(−60) ) ⇒x = ((log_2 (4−2^(−58) ))/3)

$$\mathrm{3}^{\frac{\mathrm{4}}{\mathrm{2}}+\frac{\mathrm{4}}{\mathrm{4}}+\frac{\mathrm{4}}{\mathrm{8}}+\frac{\mathrm{4}}{\mathrm{16}}+...+\frac{\mathrm{4}}{\mathrm{2}^{\mathrm{60}} }} \:=\:\mathrm{3}^{\mathrm{8}^{{x}\:} } \\ $$$$\Rightarrow\:\mathrm{4}\:\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}+...+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{60}} }\right)\:=\:\mathrm{8}^{{x}} \\ $$$$\Rightarrow\mathrm{4}\left(\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{60}} \right)}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}\right)\:=\:\mathrm{8}^{{x}} \\ $$$$\Rightarrow\mathrm{1}−\mathrm{2}^{−\mathrm{60}} \:=\:\mathrm{2}^{\mathrm{3}{x}−\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{3}{x}−\mathrm{2}\:=\:\mathrm{log}_{\mathrm{2}} \:\left(\mathrm{1}−\mathrm{2}^{−\mathrm{60}} \right) \\ $$$$\Rightarrow{x}\:=\:\frac{\mathrm{log}_{\mathrm{2}} \:\left(\mathrm{4}−\mathrm{2}^{−\mathrm{58}} \right)}{\mathrm{3}} \\ $$

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