Simple indices

Algebra Level 4

$\LARGE \sqrt[x^4]{4^x} = 4^{\sqrt[4x]{x^4}}$

If $x\ne \pm 1, 0$ is a real number satisfying the equation above, find $x$ .

Give your answer to 2 decimal places.

This is one part of the set Fun with exponents

The answer is -0.33.

2 solutions

Ashish Menon
May 3, 2016

$\begin{aligned} \LARGE \sqrt[x^4]{4^x} & = \LARGE 4^{\sqrt[4x]{x^4}}\\ \\ \LARGE {\left(4^x\right)}^{\tfrac{1}{x^4}} & = \LARGE 4^{{\left(x^4\right)}^{\frac{1}{4x}}}\\ \\ \LARGE 4^{\tfrac{1}{x^3}} & = \LARGE 4^{x^{\tfrac{1}{x}}}\\ \\ \text{Equating the powers}:-\\ \Large x^{-3} & = \Large x^{\tfrac{1}{x}}\\ \\ \text{Equating the powers}:-\\ \Large -3 & = \Large \dfrac{1}{x}\\ \\ \Large x & = \Large \dfrac{1}{-3}\\ \\ & = \Large \boxed{-0.33} \end{aligned}$

Didnt expect so easy for lvl 4..

Pawan pal - 5 years, 1 month ago

Abhay Tiwari
May 3, 2016

$\large 4^{x\times\frac{1}{x^{4}}}=4^{x^{4\times\frac{1}{4x}}}$

$\large x\times\frac{1}{x^{4}}=x^{4\times\frac{1}{4x}}$

$\large x^{-3}=x^{\frac{1}{x}}$

$-3=\frac{1}{x}$

$x=\boxed{-0.33}$

Solved it the same way :)

Ashish Menon - 5 years, 1 month ago

Yes, we think in the same way I guess? or maybe this question can be solved in this way only. :)

Abhay Tiwari - 5 years, 1 month ago

I think there can be multiple ways of solving this question. :V

Ashish Menon - 5 years, 1 month ago

@Ashish Menon – Ohk, then my former statement is correct. ;)

Abhay Tiwari - 5 years, 1 month ago

@Abhay Tiwari – Haha, true.

Ashish Menon - 5 years, 1 month ago

@Nitesh Chaudhary @Ashish Siva @Abhay Tiwari Please take your conversation somewhere else.This is a discussion forum and not a place to debate on irrelevant topics. Thanks.

Nihar Mahajan - 5 years, 1 month ago

Simple indices

This is one part of the set Fun with exponents

The answer is -0.33.

2 solutions

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