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Question Number 137585 by bramlexs22 last updated on 04/Apr/21

$$Findthecubeofthenumber$$$$N=\sqrt{7\sqrt{3\sqrt{7\sqrt{3\sqrt{7\sqrt{3\sqrt{7\sqrt{3...}}}}}}}}$$

Answered by bemath last updated on 04/Apr/21

$$N=7^{\frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\frac{1}{128}+...}.3^{\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...}$$$$N=7^{\frac{1/2}{1-1/4}}.3^{\frac{1/4}{1-1/4}}$$$$N=7^{\frac{2}{3}}.3^{\frac{1}{3}}\iff N^3=49\times 3$$$$N^3=147$$

Answered by mr W last updated on 04/Apr/21

$$N=\sqrt{7\sqrt{3N}}$$$$N^2=7\sqrt{3N}$$$$N^4=7^2\times 3N$$$$\Rightarrow N^3=7^2\times 3=147$$

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