When exponents get funny

Algebra Level 5

$\huge {\sqrt[4]{\sqrt[x]{4}}}^{\tfrac{1}{\sqrt[4]{{\left(\sqrt[x]{4x^4}\right)}^{x^4}}}} = 4^{4^{-17}x^{-x^3 - 1}}$

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

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

Want to have Fun with exponents ?

The answer is 4.00.

1 solution

Ashish Menon
May 13, 2016

$\begin{aligned} \huge {\sqrt[4]{\sqrt[x]{4}}}^{\tfrac{1}{\sqrt[4]{{\left(\sqrt[x]{4x^4}\right)}^{x^4}}}} & = \huge 4^{4^{-17}x^{-x^3 - 1}}\\ \\ \huge {\left({\left({\left(4\right)}^{\tfrac{1}{x}}\right)}^{\frac{1}{4}}\right)}^{\frac{1}{{\left({\left({\left(4x^4\right)}^{\frac{1}{x}}\right)}^{x^4}\right)}^{\frac{1}{4}}}} & = \huge 4^{4^{-17}x^{-x^3 - 1}}\\ \\ \huge {\left(4^{\tfrac{1}{4x}}\right)}^{\frac{1}{{\left(4x^4\right)}^{\frac{x^3}{4}}}} & = \huge 4^{4^{-17} × x^{-x^3 - 1}}\\ \\ \huge {\left(4^{\tfrac{1}{4x}}\right)}^{\frac{1}{{4}^{\frac{x^3}{4}} × x^{x^3}}} & = \huge 4^{4^{-17} × x^{-x^3 - 1}}\\ \\ \huge 4^{4^{-1} × x^{-1} × 4^{\frac{-x^3}{4}} × x^{x^3}} & = \huge 4^{4^{-17} × x^{-x^3 - 1}}\\ \\ \huge 4^{4^{\frac{-x^3}{4} - 1} × x^{x^3 - 1}} & = \huge 4^{4^{-17} × x^{-x^3 - 1}}\\ \\ \text{Taking root of} x^{x^3 - 1} \text{on both sides}:-\\ \huge 4^{4^{\frac{-x^3 - 4}{4}}} & = \huge 4^{4^{-17}}\\ \\ \text{Equating the powers, we get}:-\\ \huge 4^{\tfrac{-x^3 - 4}{4}} & = \huge 4^{-17}\\ \\ \text{Equating the powers, we get}:-\\ \Large \dfrac{-x^3 - 4}{4} & = \Large -17\\ \\ \Large -x^3 - 4 & = \Large -68\\ \\ \Large -x^3 & = \Large -64\\ \Large x^3 & = \Large 64\\ \Large x & =\Large \boxed{4} \end{aligned}$

Must be hard for you to come out with this question🤣

donglin loo - 3 years ago

This doodle is nothing but a confusing cramp!!!

Andreas Wendler - 5 years, 1 month ago

Sorry I was not able to write it as square root, it was not working, dont know why, maybe being poliite provides encouragement. I have worked very hard on this solution. Writing long LaTeX is not that simple. We have to take extra care of { and } Plz toggle this LaTeX to see how long it is, so a comment like this discourages a contribute to contribute more.

Ashish Menon - 5 years, 1 month ago

I know how you feel, bro. The fact that you had the patience to write this is actually something worth complimenting. Anyway...

Correction time!

Row 5: $\Large x^{-x^3}$

Row 6: $\Large x^{-x^3-1}$

Below there: Take the root of $x^{-x^3-1}$

Anyway, I don't recommend taking the root away like this. I recommend factorizing it and showing that there is no solution for that factor. Since it's too troublesome to edit so much, I'll write it here:

$\Large4^{4^{\frac{-x^3-4}{4}} \times x^{-x^3-1}} = 4^{4^{-17} \times x^{-x^3-1}}$

The base is the same, so we equate the powers:

$\Large4^{\frac{-x^3-4}{4}} \times x^{-x^3-1}=4^{-17} \times x^{-x^3-1}\\ \Large 4^{\frac{-x^3-4}{4}} \times x^{-x^3-1} - 4^{-17} \times x^{-x^3-1} = 0\\ \Large x^{-x^3-1}\left(4^{\frac{-x^3-4}{4}} -4^{-17} \right) = 0\\ \Large x^{-x^3-1} = 0,\quad 4^{\frac{-x^3-4}{4}} -4^{-17} =0$

The solution in the right factor is displayed in the original solution. I will show that the left factor has no solution.

We know that for any $a\neq 0,\;a^b\neq0$ for all values of $b$

Therefore, for any $x\neq 0,\; x^{-x^3-1} \neq 0$

$x$ cannot be $0$ according to the question, therefore there is no solution for this factor

Hung Woei Neoh - 5 years, 1 month ago

@Hung Woei Neoh – Oh thanks ^_^

Ashish Menon - 5 years, 1 month ago

Exactly it's very very difficult

Arnav Das - 4 years, 11 months ago

@Arnav Das – Yep, :) :) thanks for supporting!

Ashish Menon - 4 years, 11 months ago

When exponents get funny

Want to have Fun with exponents ?

The answer is 4.00.

1 solution

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