An algebra problem by John Rex Alegria

Algebra Level 2

x 6 ! 3 ! 5 ! = [ 2 x x + 15 x 2 x 198 x ] 2 y \large \sqrt[5!]{\sqrt[3!]{x^{6!}}} = \left [ \dfrac{2x^x + \sqrt[x]{15x^2}}{198x} \right ] ^{2y}

If x > 0 x> 0 and y = 0 y = 0 , find the value of ( x − 1 ) x 2 (x-1)^{x^2} .

Notation : ! ! denotes the factorial notation. For example, 8 ! = 1 × 2 × 3 × ⋯ × 8 8! = 1\times2\times3\times\cdots\times8 .


The answer is 0.

John Rex Alegria by John Rex Alegria , 23, Philippines

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2 solutions

x 6 ! 3 ! 5 ! = [ ⋯ ] 2 y x 6 ! 3 ! 5 ! = [ ⋯ ] 0 x 6 ! 3 ! 5 ! = 1 x = 1 \begin{aligned} \sqrt [5!]{\sqrt [3!]{x^{6!}}} & = [\cdots]^{2y} \\ \sqrt [5!]{x^{\frac {6!}{3!}}} & = [\cdots]^{0} \\ x^{\frac {6!}{3!5!}} & = 1 \\ x&=1 \end{aligned}

⟹ ( x − 1 ) x 2 = 0 \implies (x-1)^{x^2} =\boxed {0}

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The task seems a bit senseless and hairsplitting I guess!

Andreas Wendler - 4 years, 9 months ago

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Yes, not a good problem.

Chew-Seong Cheong - 4 years, 9 months ago
Sswag SSwagf
Sep 23, 2016

[ 2 x x + 15 x 2 x 198 x ] 2.0 = 1 = [ x 6 ! 3 ! 5 ! \left [ \dfrac{2x^x + \sqrt[x]{15x^2}}{198x} \right ] ^{2.0} =1 =[\large \sqrt[5!]{\sqrt[3!]{x^{6!}}} = x 6 ! 3 ! 5 ! = 1 x^\frac {6!}{3!5!}=1

( 1 − 1 ) x 2 = 0 x 2 (1-1)^{x^2}=0^{x^2} = 0 0

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