How much numbers In-dice

Algebra Level 3

2 x 3 x 4 x = 3 x 4 x 2 x 2\sqrt[4x]{x^{3x}} = 3\sqrt[2x]{x^{4x}}

Find the real value of x x satisfying the equation above.

The answer is of the form a b 5 \sqrt[5]{\dfrac{a}{b}} where a a and b b are positive co-prime integers, then submit the value of a + b a + b .

Note \text{Note} :- Here x 1 , 0 , 1 x \neq -1,0,1


This is one part of the set Fun with exponents


The answer is 97.

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1 solution

Ashish Menon
Apr 29, 2016

2 x 3 x 4 x = 3 x 4 x 2 x 2 ( x 3 x ) 1 4 x = 3 ( x 4 x ) 1 2 x 2 x 3 4 = 3 x 2 2 3 = x 2 x 3 4 2 3 = x 2 3 4 2 3 = x 5 4 Raising power of 4 on both sides : 16 81 = x 5 Raising root of 5 on both sides : x = 16 81 5 a + b = 16 + 81 = 97 \begin{aligned} 2\sqrt[4x]{x^{3x}} & = 3\sqrt[2x]{x^{4x}}\\ 2{\left(x^{3x}\right)}^{\tfrac{1}{4x}} & = 3{\left(x^{4x}\right)}^{\tfrac{1}{2x}}\\ 2x^{\tfrac{3}{4}} & = 3x^2\\ \dfrac{2}{3} & = \dfrac{x^2}{x^{\tfrac{3}{4}}}\\ \dfrac{2}{3} & = x^{2 - \tfrac{3}{4}}\\ \dfrac{2}{3} & = x^{\tfrac{5}{4}}\\ \text{Raising power of 4 on both sides}:-\\ \dfrac{16}{81} & = x^5\\ \text{Raising root of 5 on both sides}:-\\ x & = \sqrt[5]{\dfrac{16}{81}}\\ \therefore a + b & = 16 + 81\\ & = \boxed{97} \end{aligned}

It's a/b right? U wrote ab in the question.

Abhiram Rao - 5 years, 1 month ago

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Oh yes thanks!I have changed it. Problem with LaTeX.

Ashish Menon - 5 years, 1 month ago

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A week to go!

Abhiram Rao - 5 years, 1 month ago

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@Abhiram Rao For what !?!?!?

Ashish Menon - 5 years, 1 month ago

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@Ashish Menon Cap America: Civil War

Abhiram Rao - 5 years, 1 month ago

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@Abhiram Rao Watching Thor. Ttyl

Abhiram Rao - 5 years, 1 month ago

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@Abhiram Rao Ttyl gn ...

Ashish Menon - 5 years, 1 month ago

@Abhiram Rao Oh ;) ;) great

Ashish Menon - 5 years, 1 month ago

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