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Algebra Level 3

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

Given that x ( ≠ 1 , 0 , − 1 ) x\, (\ne 1,0,-1) satisfies the equation above and x x can be expressed as a b , \frac ab, where a a and b b are coprime positive integers, find a + b . a+b.

This is one part of the set Fun with exponents .


The answer is 881.

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Ashish Menon by Ashish Menon , 20, India

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4 solutions

Ashish Menon
Apr 12, 2016

( x x 4 ) x = x x x 4 x x . x 1 4 x = x x × x 1 4 x x + 1 4 x = x x 1 + 1 4 x 5 4 x = x x 5 4 Equating the powers 5 4 x = x 5 4 5 4 = x 5 4 x 5 4 = x 5 4 − 1 5 4 = x 1 4 Raising 4 on both sides ( 5 4 ) 4 = x x = 625 256 ∴ a + b = 625 + 256 = 881 \begin{aligned} {\left(x\sqrt[4]{x}\right)}^x & = x^{x\sqrt[4]{x}}\\ x^x.x^{\tfrac{1}{4}x} & = x^{x × x^{\tfrac{1}{4}}}\\ x^{x + \tfrac{1}{4}x} & = x^{x^{1 + \tfrac{1}{4}}}\\ x^{\tfrac{5}{4}x} & = x^{x^{\tfrac{5}{4}}}\\ \text{Equating the powers}\\ \dfrac{5}{4}x & = x^{\tfrac{5}{4}}\\ \dfrac{5}{4} & = \dfrac{x^{\tfrac{5}{4}}}{x}\\ \dfrac {5}{4} & = x^{\tfrac{5}{4} - 1}\\ \dfrac {5}{4} & = x^{\tfrac{1}{4}}\\ \text{Raising 4 on both sides}\\ {(\dfrac{5}{4})}^4 & = x\\ x & = \dfrac{625}{256}\\ \therefore a + b & = 625 + 256\\ & = \boxed{881} \end{aligned}

15 Helpful 4 Interesting 1 Brilliant 1 Confused

Moderator note:

This solution isn't complete. You can only "Equate the powers" if you have shown that the base is not 1, 0 or -1.

While x = 1 x=1 is ruled out (eventually) in the question, you have to rule out the possibility of x = 0 , − 1 x = 0, -1 .

@Ashish Siva Good solution! This is a potential Featured problem. Nice work! It's good to see you improving with LaTeX too!

Mehul Arora - 5 years, 2 months ago

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Thank you mehul for your words.

Ashish Menon - 5 years, 2 months ago

But pay attention that x=1 is ALSO a solution! ;-)

Andreas Wendler - 5 years, 2 months ago

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Thanks, I have edited it accordingly.

Ashish Menon - 5 years, 2 months ago

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Where is it??

Andreas Wendler - 5 years, 2 months ago

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@Andreas Wendler – That x is not equal to 1,0 part.

Ashish Menon - 5 years, 2 months ago

I got this one wrong :(

Abhiram Rao - 5 years, 1 month ago

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Then you would get next one right :)

Ashish Menon - 5 years, 1 month ago

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xD I did. 3+2+4=9

Abhiram Rao - 5 years, 1 month ago

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@Abhiram Rao – Very good :+1:

Ashish Menon - 5 years, 1 month ago

⇒ ( x x 4 ) x = x x x 4 \Rightarrow {{(\color{#20A900}{x}\color{#D61F06}{\sqrt[4]{x}})}^\color{#20A900}{x} = \color{#20A900}{x}^{\color{#20A900}{x}\color{#D61F06}{\sqrt[4]{x}}}}

( x x 4 ) = x x 4 {{(\color{#20A900}{x}\color{#D61F06}{\sqrt[4]{x}})}= \color{#20A900}{x}^{\color{#D61F06}{\sqrt[4]{x}}}}

Let x 4 = y \color{#D61F06}{\sqrt[4]{x}}=y .

x y = x y \color{#20A900}{ x}y=\color{#20A900}{ x}^{y}

y = x y x = x y − 1 y =\dfrac{\color{#20A900}{x}^y}{\color{#20A900}{x}}=\color{#20A900}{x}^{y-1}

x 4 = x x 4 − 1 \color{#D61F06}{\sqrt[4]{x}}=\color{#20A900}{x}^{\color{#D61F06}{\sqrt[4]{x}}-1}

x = x 4 x 4 − 4 \color{#20A900}{x}=\color{#20A900}{x}^{\color{#D61F06}{4\sqrt[4]{x}}-4}

1 = 4 x 4 − 4 1=\color{#EC7300}{4}\color{#D61F06}{\sqrt[4]{x}}-\color{#EC7300}{4}

x 4 = 5 4 \color{#D61F06}{\sqrt[4]{x}}=\dfrac{\color{#3D99F6}{5}}{\color{#EC7300}{4}}

x = ( 5 4 ) 4 \color{#20A900}{x}=\left(\dfrac{\color{#3D99F6}{5}}{\color{#EC7300}{4}}\right)^\color{#EC7300}{4}

x = 625 256 = a b \color{#20A900}{x}=\dfrac{\color{#3D99F6}{625}}{\color{#EC7300}{256}}=\dfrac{\color{#3D99F6}{a}}{\color{#EC7300}{b}}

∴ a + b = 625 + 256 = 881 \color{#E81990}{\therefore} \color{#3D99F6}{a}+\color{#EC7300}{b}=\color{#3D99F6}{625}+\color{#EC7300}{256}=\boxed{\color{#624F41}{881}}

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Jonathan Hocker
Apr 12, 2016

Isn't x=1 also a solution? Which would mean that 1=a/b so a+b=2a=any positive real (due to the even root of x and not allowing 0^0).

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Thanks I have edited it accordingly.

Ashish Menon - 5 years, 2 months ago

In future, if you spot any errors with a problem, you can “report” it by selecting "report problem" from the menu. This will notify the problem creator who can fix the issues.

Calvin Lin Staff - 5 years, 2 months ago
Atul Shivam
Apr 12, 2016

Given the equation ( x x 4 ) x = x ( x x 4 ) (x\sqrt[4]{x})^x=x^{(x\sqrt[4]{x})}

Which can be written as ( x ( 5 4 ) x ) = x ( x 5 4 ) (x^{(\frac{5}{4})^x})=x^{(x^{\frac{5}{4}})}

( x ) 5 4 x = x ( x 5 4 ) (x)^{\frac{5}{4}x}=x^{(x^{\frac{5}{4}})}

Now Taking l o g log on both sides we have 5 4 x l o g x = ( x 5 4 ) l o g x \frac{5}{4}xlog x=(x^{\frac{5}{4}})log x

Now equation can be reduced to 5 4 x = ( x 5 4 ) \frac{5}{4}x=(x^{\frac{5}{4}})

x 5 4 x = 5 4 \frac{x^{\frac{5}{4}}}{x}=\frac{5}{4}

x 1 4 = 5 4 x^{\frac{1}{4}}=\frac{5}{4}

x = ( 5 4 ) 4 = 625 256 = a b x=(\frac{5}{4})^4=\frac{625}{256}=\frac{a}{b}

a + b = 881 a+b=\boxed{881}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Always check the conditions of the "theorem" that you are applying.

Be careful when applying logs. It might not work out great if x < 0 x < 0 , x = 0 x = 0 or x = 1 x = 1 .

A slight mistake sir, you should correct 4 to 4^4 in the last lines.

Silver Vice - 5 years, 2 months ago

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Thanks I've corrected it!!! :-)

Atul Shivam - 5 years, 2 months ago

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