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Question

I m not able to find sign of square root in the keyboard. I want to know if 7 is under square root of 7 many times. How to solve such questions?

Answer 1

Let y=√(7√(7√7)), then y^2=7√(7√7); y^4=7^2*7√7=7^3√7; y^8=7^6*7=7^7.

So y=7^(7/8)=5.4886 approx. In general, if N is the number of 7's under the square root then y=7^(((2^N)-1)/2^N). In this case N=3 so y=7^((2^3 - 1)/2^3)=7^(7/8). For N=4, y=7^(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=√(7(√7(√7... So y^2=7√7(√7√7... =7y, so since y≠0, y=7 (dividing through by y).