Complex Exponents

Algebra Level 5

( ( x ) 2 x ) 3 x x i 2 = ( ( ( x ) 2 x ) 3 x ) 4 x 1 5 x \huge \sqrt[{x^i}^2]{{\left({\left(x\right)}^{2x}\right)}^{3x}} = \sqrt[\frac{1}{5x}]{{\left({\left({\left(x\right)}^{2x}\right)}^{3x}\right)}^{4x}}

Find the real value of x x (upto 2 decimal places) satisfying the real equation above.

Note :- Here x ≠ { − 1 , 0 , 1 } x \neq \{-1 , 0 , 1\} . Here i i denotes − 1 \sqrt{-1} .

This is one part of the set Fun with exponents


The answer is 0.05.

Ashish Menon by Ashish Menon , 20, India

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4 solutions

Ashish Menon
May 6, 2016

( ( x ) 2 x ) 3 x x i 2 = ( ( ( x ) 2 x ) 3 x ) 4 x 1 5 x ( x ) 6 x 2 x − 1 = ( ( ( ( x ) 2 x ) 3 x ) 4 x ) 5 x x 6 x 3 = x 120 x 4 Equating the powers : − 6 x 3 = 120 x 4 x = 6 120 = 0.05 \begin{aligned} \huge \sqrt[{x^i}^2]{{\left({\left(x\right)}^{2x}\right)}^{3x}} & = \huge \sqrt[\frac{1}{5x}]{{\left({\left({\left(x\right)}^{2x}\right)}^{3x}\right)}^{4x}}\\ \\ \huge \sqrt[x^{-1}]{{\left(x\right)}^{6x^2}} & = \huge {\left ({\left({\left({\left(x\right)}^{2x}\right)}^{3x}\right)}^{4x}\right)}^{5x}\\ \\ \LARGE x^{6x^3} & = \LARGE x^{120x^4}\\ \\ \text{Equating the powers}:-\\ \Large 6x^3 & = \Large 120x^4\\ \\ \Large x & = \dfrac{6}{120}\\ & = \boxed{0.05} \end{aligned}

8 Helpful 0 Interesting 0 Brilliant 0 Confused

⇒ ( ( x ) 2 x ) 3 x x i 2 = ( ( ( x ) 2 x ) 3 x ) 4 x 1 5 x \Rightarrow \sqrt[x^{\color{#D61F06}{i^2}}]{{\left({\left(x\right)}^{2x}\right)}^{3x}} = \sqrt[\frac{1}{5x}]{{\left({\left({\left(x\right)}^{2x}\right)}^{3x}\right)}^{4x}}

x 6 x 2 x − 1 = x 24 x 3 × 5 x x^{\frac{6x^2}{x^{\color{#D61F06}{-1}}}}=x^{24x^3×5x}

6 x 3 = 24 x 3 × 5 x 6x^3=24x^3×5x

x = 1 20 = 0.05 x=\dfrac{1}{20}=\boxed{0.05}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Yes, it is the exact same method as mine XD

Ashish Menon - 5 years, 1 month ago

I don't understand why is this level 4 algebra, suppose to be level 3

Jason Chrysoprase - 5 years, 1 month ago
Sabhrant Sachan
May 6, 2016

Your Questions on Exponents are of the Highest Quality currently in Brilliant \text {Your Questions on Exponents are of the Highest Quality currently in Brilliant }

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Jeez thanks! :)

Ashish Menon - 5 years, 1 month ago

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I told you :P

Abhiram Rao - 5 years, 1 month ago

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Thanks brother :)

Ashish Menon - 5 years, 1 month ago

i 2 = − 1 , b a = b a − 1 , a n d ( X m ) n = X m × n . Since the base on both sides is same we we equate the powers as follows:- ( 1 X − 1 ) ( 3 X ) ( 2 X ) = ( 1 ( 5 X ) − 1 ) ( 4 X ) ( 3 X ) ( 2 X ) ⟹ ( X ) = ( 5 X ) ( 4 X ) , ⟹ X = 1 20 = 0.05. \ \ \ \large i^2= - 1,\ \ \ \sqrt[a]b=b^{a^{ -1 }},\ \ \ and\ \ (X^m)^n=X^{m \times n}.\ \ \\ \text{Since the base on both sides is same we we equate the powers as follows:-}\\ (\dfrac 1 {X^{ - 1}})(3X)(2X)\ =\ (\dfrac 1 {(5X)^{ - 1}})(4X)(3X)(2X)\\ \implies\ (X)=(5X)(4X),\ \ \ \implies X=\frac 1 {20}=0.05.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Correct! One suggestion:- The solution might look more stylish if you replace X X with x x ;)

Ashish Menon - 5 years, 1 month ago

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