Crazy exponents

Algebra Level 4

( ( x ) 4 4 x 16 ) 16 x 8 4 = ( ( x 4 x 2 ) 4 x 8 8 ) 16 x \Large {\left({\left(\sqrt{x}\right)}^{4\sqrt{4x^{16}}}\right)}^{\sqrt[4]{16x^8}} = {\left({\left(x^{4x^2}\right)}^{\sqrt[8]{4x^8}}\right)}^{16\sqrt{\sqrt{x}}}

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

The answer is of the form a 27 \sqrt[27]{a} , then submit the value of a a

Here x ≠ 0 , 1 x \neq 0,1

This is one part of the set Fun with exponents


The answer is 8192.

Ashish Menon by Ashish Menon , 20, India

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1 solution

Ashish Menon
Apr 26, 2016

( ( x ) 4 4 x 16 ) 16 x 8 4 = ( ( x 4 x 2 ) 4 x 8 8 ) 16 x ( ( x 1 2 ) 8 x 8 ) 2 x 2 = ( ( x 4 x 2 ) 4 x 8 8 ) 16 x 1 4 x 1 2 × 8 x 8 × 2 x 2 = x 4 x 2 × 4 x 8 8 × 6 x 1 4 x 8 x 10 = x 64 4 8 x 13 4 Equating the powers : − 8 x 10 = 64 4 8 x 13 4 8 64 4 8 = x 13 4 x 10 x 10 x 13 4 = 8 4 8 x 27 4 = 8 4 8 Raising power of 4 on both sides : − x 27 = 8192 x = 8192 27 ∴ a = 8192 \begin{aligned} \Large{{\left({\left(\sqrt{x}\right)}^{4\sqrt{4x^{16}}}\right)}^{\sqrt[4]{16x^8}}} & = \Large{{\left({\left(x^{4x^2}\right)}^{\sqrt[8]{4x^8}}\right)}^{16\sqrt{\sqrt{x}}}}\\ \\ \Large{{\left({\left(x^{\tfrac{1}{2}}\right)}^{8x^8}\right)}^{2x^2}} & = \Large{{\left({\left(x^{4x^2}\right)}^{\sqrt[8]{4x^8}}\right)}^{16x^{\frac{1}{4}}}}\\ \\ \Large{x^{\tfrac{1}{2} × 8x^8 × 2x^2}} & = \Large{x^{4x^2 × \sqrt[8]{4x^8} × 6x^{\frac{1}{4}}}}\\ \\ \Large{x^{8x^{10}}} & = \Large{x^{64\sqrt[8]{4}x^{\frac{13}{4}}}}\\ \\ \text{Equating the powers}:-\\ \Large{8x^{10}} & = \Large{64\sqrt[8]{4} x^{\tfrac{13}{4}}}\\ \Large{\dfrac {8}{64\sqrt[8]{4}}} & = \Large{\dfrac{x^{\tfrac{13}{4}}}{x^{10}}}\\ \\ \Large{\dfrac{x^{10}}{x^{\frac{13}{4}}}} & = \Large{8\sqrt[8]{4}}\\ \\ \Large{x^{\frac{27}{4}}} & = \Large{8\sqrt[8]{4}}\\ \text{Raising power of 4 on both sides}:-\\ \Large{x^{27}} & = \Large{8192}\\ \Large{x} & = \Large{\sqrt[27]{8192}}\\ \therefore \Large{a} & = \Large{\boxed{8192}} \end{aligned}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

How is it that there is no space for writing solution? Any way I am giving it below.
Since the base is same I am directly equating the exponents on both sides as per rule ( x m ) n = x m × n . (x^m)^n=x^{m\times n}.
( 1 2 ) × ( 4 × 2 X 8 ) × ( 2 X 2 ) = ( 4 X 2 ) × ( 4 1 8 X ) × ( 16 × X 1 4 ) ⟹ 2 − 1 × 8 × 2 × X 8 × X 2 = 4 × 2 1 4 × 16 × X 2 × X × X 1 4 ⟹ 2 − 1 + 3 + 1 × X 8 + 2 = 2 2 + 1 4 + 4 × X 2 + 1 + 1 4 . ⟹ 2 3 × X 10 = 2 25 4 × X 13 4 ⟹ X 27 4 = 2 13 4 ⟹ X = 2 13 27 = a 27 . ⟹ a = 8192 \Large (\frac 1 2) \times (4 \times 2X^8)\times (2X^2) = (4X^2)\times (4^{\frac 1 8}X) \times(16 \times X^{\frac 1 4})\\ \Large \implies 2^{-1} \times 8 \times 2\ \ \times X^8 \times X^2 = 4 \times 2^{\frac 1 4} \times 16\ \ \times X^2 \times X \times X^{\frac 1 4}\\ \Large \implies 2^{ - 1+3+1} \times X^{8+2}=2^{2+ {\frac 1 4} +4}\ \ \times X^{2+1+ {\frac 1 4}}. \\ \Large \implies 2^3\times X^{10} = 2^{\frac {25} 4} \times X^{ \frac {13} 4}\\ \Large \implies X^{\frac {27} 4}= 2^{\frac {13} 4} \\ \Large \implies X=\sqrt[27]{2^{13}}=\sqrt[27]a.\\ \Large \implies a = \color{#D61F06}{8192}

Niranjan Khanderia - 5 years, 1 month ago

Ehh I'm confused. How did you get this:

8 x 10 = 64 4 8 − 8 x 10 8x^{10} = 64 \sqrt[8]{4} - 8x^{10}

Hung Woei Neoh - 5 years, 1 month ago

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You were not confused, my solution was wrong. The answer is correct, but I typed the same LaTeX two times XD. Thank you very much, I have edited my solution accordingly.

Ashish Menon - 5 years, 1 month ago

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You're welcome :)

Hung Woei Neoh - 5 years, 1 month ago

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