Roots inside roots?

Algebra Level 4

$\huge{\sqrt[{\sqrt[4]{4x}}]{{(x^4)}^{\sqrt[4]{x}}} = x^x}$

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

The answer is of the form $a × \sqrt[4]{b}$ , where $a$ and $b$ are integers, then what is $a + b$ ?

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

The answer is 6.

2 solutions

Ashish Menon
Apr 30, 2016

$\begin{aligned} \huge\sqrt[{\sqrt[4]{4x}}]{{\left(x^4\right)}^{\sqrt[4]{x}}} & = \huge x^x\\ \\ \huge \sqrt[{(4x)}^{\tfrac{1}{4}}]{x^{4 × x^{\tfrac{1}{4}}}} & = \huge x^x\\ \\ \huge x^{4 × x^{\tfrac{1}{4}} × \tfrac{1}{4^{\frac{1}{4}} × x^{\frac{1}{4}}}} & = \huge x^x\\ \\ \huge x^{\tfrac{4}{4^{\frac{1}{4}}}} & = \huge x^x\\ \\ \huge x^{4^{1 - \tfrac{1}{4}}} & = \huge x^x\\ \\ \text{Equating the powers}:-\\ \LARGE 4^{\tfrac{3}{4}} & = \LARGE x\\ \\ \LARGE x & = \LARGE \sqrt[4]{64}\\ \\ \LARGE x & = \LARGE \sqrt[4]{16 × 4}\\ \\ \LARGE x & = \LARGE 2\sqrt[4]{4}\\ \\ \therefore \LARGE a + b & = \LARGE 2+ 4\\ & = \LARGE \boxed{6} \end{aligned}$

I confess, you are making some serious problems now. :). This is one of your Great Problem on roots. Learning a lot from your problems.

Abhay Tiwari - 5 years, 1 month ago

Jeez, thanks!

Ashish Menon - 5 years, 1 month ago

Please mention that a and b are integers, because $a=4$ and $b=\frac{1}{4}$ are also correct. Otherwise, nice problem!

Nanda Rahsyad - 5 years, 1 month ago

Thanks for your comment but I am curious to know how b=1/4 though I am mentioning it in my question :) Thanks again!

Ashish Menon - 5 years, 1 month ago

Oops, my bad... I meant $a=4$ ... sorry :)

Nanda Rahsyad - 5 years, 1 month ago

@Nanda Rahsyad – Ah! I get your point thanks.

Ashish Menon - 5 years, 1 month ago

Correction: Row 5

$\huge x^{4^{1-\frac{1}{4}}} = x^x$

Hung Woei Neoh - 5 years, 1 month ago

Thanks :) :+1:

Ashish Menon - 5 years, 1 month ago

Niranjan Khanderia
May 5, 2016

$\Large { \sqrt[ { \sqrt[4] {4X} } ]{ { (X^4)}^{ \sqrt[4]{X} } } \\ = {\huge ( X^4)}^{^{{\normalsize \dfrac {X ^{\frac 1 4}}{4^{\frac 1 4}*X^{\frac 1 4}} } }} \\ ={\huge ( X^4)}^{^{{\normalsize \dfrac 1 {4^{\frac 1 4}}}} } \\ ={\huge X}^{^{{\normalsize \dfrac 4{4^{\frac 1 4}} } }} \\ =X^{2\sqrt2} = X^X \\ \text{Base being common, equating the powers}\\ X=2\sqrt2=2\sqrt[4]{4} =a\sqrt[4]{b} \\ a+b=6 }$

Exactly. The solution would look better if you use the LaTeX code \times in place of * :)

Ashish Menon - 5 years, 1 month ago

Thanks. I will do that. It is only one place here! How about my raise to ! I have used double ^^.

Niranjan Khanderia - 5 years, 1 month ago

Yes, :P great :+1:

Ashish Menon - 5 years, 1 month ago

@Ashish Menon – Thank you.

Niranjan Khanderia - 5 years, 1 month ago

@Niranjan Khanderia – My pleasure.

Ashish Menon - 5 years, 1 month ago