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

Find the cube of the number N= (√(7(√(3(√(7(√(3(√(7(√(3(√(7(√(3...))))))))))))))))

$${Find}\:{the}\:{cube}\:{of}\:{the}\:{number}\: \\ $$$${N}=\:\sqrt{\mathrm{7}\sqrt{\mathrm{3}\sqrt{\mathrm{7}\sqrt{\mathrm{3}\sqrt{\mathrm{7}\sqrt{\mathrm{3}\sqrt{\mathrm{7}\sqrt{\mathrm{3}...}}}}}}}} \\ $$

Answered by bemath last updated on 04/Apr/21

N=7^((1/2)+(1/8)+(1/(32))+(1/(128))+...) . 3^((1/4)+(1/(16))+(1/(64))+...) N=7^((1/2)/(1−1/4)) . 3^((1/4)/(1−1/4)) N= 7^(2/3) . 3^(1/3) ⇔ N^3 = 49×3 N^3 = 147

$${N}=\mathrm{7}^{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{32}}+\frac{\mathrm{1}}{\mathrm{128}}+...} .\:\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{16}}+\frac{\mathrm{1}}{\mathrm{64}}+...} \\ $$$${N}=\mathrm{7}^{\frac{\mathrm{1}/\mathrm{2}}{\mathrm{1}−\mathrm{1}/\mathrm{4}}} \:.\:\mathrm{3}^{\frac{\mathrm{1}/\mathrm{4}}{\mathrm{1}−\mathrm{1}/\mathrm{4}}} \\ $$$${N}=\:\mathrm{7}^{\frac{\mathrm{2}}{\mathrm{3}}} .\:\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} \:\Leftrightarrow\:{N}^{\mathrm{3}} \:=\:\mathrm{49}×\mathrm{3} \\ $$$${N}^{\mathrm{3}} \:=\:\mathrm{147} \\ $$

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

N=(√(7(√(3N)))) N^2 =7(√(3N)) N^4 =7^2 ×3N ⇒N^3 =7^2 ×3=147

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

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