Japanese | Can you solve this ? | A Nice Math Olympiad Problem with Square Roots

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[@ddichny]

X * Sqrt( X * Sqrt(X) ) = 2 Since square root is same as to the power of one half, then: X * ( X * X ^ (1/2) ) ^ (1/2) = 2 X * ( X ^ (1 + 1/2) ) ^ (1/2) = 2 [ via N^A * N^B = N^(A+B) ] X * ( X ^ (3/2) ) ^ (1/2) = 2 X * ( X ^ (3/2 * 1/2) ) = 2 [ via (N^A)^B = N^(A*B) X * ( X ^ (3/4) ) = 2 X ^ (1 + 3/4) = 2 X ^ (7/4) = 2 X = 2 ^ (4/7) ...which is same as his "7th root of 16", since 16 = 2^4 and (2^4)^(1/7) = 2^(4/7)

[@SolutionofMaths]

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[@antonellocossu4319]

Nice video. I liked the final verification a lot...

[@learncommunolizer]

Thank you very much

[@Yureka-ox5jn]

Nice équation💯💯❤🧡💛💚💙💜🖤

[@learncommunolizer]

Thank you very much ❤️❤️❤️

[@phoebe543]

Can someone explain where i went wrong, please? X^3•x^2•x = 2^3 X^6 = 2^3 X^6/6 = 2^3/6 X = sqrt(2)

[@ddichny]

The square root of X, cubed (your second term on the left) is not X^2, it's X to the power of (3/2). Square root of the square root of X, cubed (your third term on the left), is not x, it's X to the power of (3/4).

[@phoebe543]

@@ddichny, thanks. It was dumber than that, really, I wrote ^3 instead of ^4. Should be both sides squared then (X^2)^2 X^2 X = (2^2)^2 X^4X^2X = 2^4 X^7 = 2^4 X = 2^4/7 X = 7 root (2^4) X = 7root(16)