Funny exponents

Algebra Level 4

$\huge \sqrt[\large{\dfrac{1}{4x}}]{{\left(\sqrt[4x]{4^x}\right)}^{x^4}} = 4^{\left(\sqrt[4x]{{\left(\sqrt[4x^4]{x^4}\right)}^{x^4}}\right)}$

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

The answer is of the form $\dfrac{a}{b}$ where $a$ and $b$ are positive co-prime integers. Then find the value of $a + b$ .

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

This is one part of the set Fun with exponents

The answer is 21.

1 solution

Ashish Menon
Apr 29, 2016

$\begin{aligned} \huge \sqrt[\large{\dfrac{1}{4x}}]{{\left(\sqrt[4x]{4^x}\right)}^{x^4}} & = \huge 4^{\left(\sqrt[4x]{{\left(\sqrt[4x^4]{x^4}\right)}^{x^4}}\right)}\\ \\ \huge {\left({\left({\left({4^x}\right)}^{\tfrac{1}{4x}}\right)}^{x^4}\right)}^{4x} & = \huge 4^{{\left({\left({\left({x^4}\right)}^{\tfrac{1}{4x^4}}\right)}^{x^4}\right)}^{\frac{1}{4x}}}\\ \\ \huge 4^{x^5} & = \huge 4^{x^{\frac{1}{4x}}}\\ \\ \text{Equating the powers}:-\\ \huge x^5 & = \huge x^{\tfrac{1}{4x}}\\ \\ \text{Equating the powers}:-\\ \LARGE 5 & = \LARGE \dfrac{1}{4x}\\ \\ \LARGE x & = \LARGE \dfrac{1}{20}\\ \\ \LARGE \therefore a + b & = \LARGE 1 + 20\\ & = \LARGE \boxed{21} \end{aligned}$

On the LHS term of the first step the last exponent is $\frac{1}{4x}$ but in the second step you wrote the last exponent of LHS as $4x$ . Please check the question once again.

Rudraksh Shukla - 5 years, 1 month ago

I think I had modified this question 5 - 6 hrs ago.