Beginner Mathematics-8

Algebra Level 3

The value of x x satisfying x x = 2 2 x^x = \frac{\sqrt{2}}{2} may be expressed as a b , \frac{a}{b}, where gcd ⁡ ( a , b ) = 1 \gcd(a, b) = 1 and b b is not doubly even.

Find a + b . a + b.

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Ekene Franklin by EKENE FRANKLIN , 16, Nigeria

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2 solutions

x x = 2 2 = 1 2 = 1 2 = ( 1 2 ) 1 2 ⟹ x = 1 2 \begin{aligned} x^x & = \frac {\sqrt 2}2 = \frac 1{\sqrt 2} = \sqrt {\frac 12} = \left(\frac 12 \right)^\frac 12 \\ \implies x & = \frac 12 \end{aligned}

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@EKENE FRANKLIN , it is better to say " a a and b b are positive coprime integers" instead of "where gcd ⁡ ( a , b ) = 1 \gcd (a,b) = 1 , because a a and b b can be negative.

Chew-Seong Cheong - 2 years, 7 months ago

1 2 1 2 = 1 2 = 2 2 \large \frac{1}{2}^{\frac{1}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}

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