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Algebra Level 2

$\Large \big({\sqrt[3]{x}}\big)^{x^3} = {x^{3\sqrt[3]{x}}}$

Find the real value of $x$ which satisfies the equation above.

If the answer is of the form $\sqrt[8]{a}$ , submit the value of $a$ .

Note: Here $x \neq 1,0,-1$ .

The answer is 729.

1 solution

Ashish Menon
Apr 24, 2016

$\begin{aligned} {\sqrt[3]{x}}^{x^3} & = x^{{3}^{\sqrt[3]{x}}}\\ x^{{\tfrac{1}{3}}^{x^3}} & = x^{{3}^{x^{\frac{1}{3}}}}\\ x^{\tfrac{1}{3}x^3} & = x^{3x^{\frac{1}{3}}}\\ \text{Equating the powers}:-\\ \dfrac13x^3 & = 3x^{\tfrac13}\\ \dfrac19 & = \dfrac{x^{\tfrac13}}{x^3}\\ \dfrac19 & = x^{\tfrac13 - 3}\\ \dfrac19 & = x^{-\tfrac83}\\ \text{Cubing on both sides}:-\\ \dfrac{1}{729} & = x^{-8}\\ x^8 & = 729\\ x & = \sqrt[8]{729} \end{aligned}$

$\therefore a = \boxed{729}$

Technically, you have to show that $x = -\sqrt[8]{729}$ is not a valid solution.

Arjen Vreugdenhil - 5 years, 1 month ago

I thought that was understood :/ because the equation has to be real.

Ashish Menon - 5 years, 1 month ago

For negative $x$ , the power $x^y$ is defined and real if $y = p/q$ is rational with $q$ odd. For instance, $(-8)^{-2/3} = -\tfrac 14$ .

There is no negative solution because $x$ is not rational.

Arjen Vreugdenhil - 5 years, 1 month ago

I think you should make a set on such problems. Would be fun solving 'em all cause I only found some 2

Abhiram Rao - 5 years, 1 month ago

@Abhiram Rao – Alright :) Will do

Ashish Menon - 5 years, 1 month ago

@Abhiram Rao – Here it is :- Have fun with exponents!!!

Ashish Menon - 5 years, 1 month ago

@Ashish Menon – :D thanks!

Abhiram Rao - 5 years, 1 month ago

@Abhiram Rao – My pleasure.

Ashish Menon - 5 years, 1 month ago

Indices

The answer is 729.

1 solution

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