Lets asume: y=777, y2=777, y4=72⋅77=737, y8=76⋅7=77 So y=778=5.4886 . In general, if N is the number of 7's under the square root then y=7(2N)−12N. In this case N=3 so y=723−123=778. For N=4, y=71516=6.1984, as N approaches infinity y approaches 7. So for 7's repeated under square roots indefinitely the answer is 7. Another way of solving it is to put y=7(7(7… So y2=77(77…=7y, so since y≠0,y=7 (dividing through by y).

Explained: "What is 777 ?"

High School
Algebra
Algebra I
Irrational numbers

Khaleesi Herbert

Answered question

2020-10-27

What is

\sqrt{7 \sqrt{7 \sqrt{7}}}

Answer & Explanation

pivonie8

Skilled2020-10-28Added 91 answers

Lets asume:
$y = \sqrt{7 \sqrt{7 \sqrt{7}}}$ ,
$y^{2} = 7 \sqrt{7 \sqrt{7}}$ ,
$y^{4} = 7^{2} \cdot 7 \sqrt{7} = 7^{3} \sqrt{7}$ ,
$y^{8} = 7^{6} \cdot 7 = 7^{7}$
So $y = 7^{\frac{7}{8}} = 5.4886$ . In general, if N is the number of 7's under the square root then $y = 7^{\frac{(2^{N}) - 1}{2^{N}}}$ . In this case N=3 so $y = 7^{\frac{2^{3} - 1}{2^{3}}} = 7^{\frac{7}{8}}$ . For N=4, $y = 7^{\frac{15}{16}} = 6.1984$ , as N approaches infinity y approaches 7. So for 7's repeated under square roots indefinitely the answer is 7.
Another way of solving it is to put $y = \sqrt{7 (\sqrt{7} (\sqrt{7} \dots}$ So $y^{2} = 7 \sqrt{7} (\sqrt{7} \sqrt{7} \dots = 7 y$ , so since $y \neq 0, y = 7$ (dividing through by y).

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