How powerful is power?

Algebra Level 3

If x x = 2015 , \Large {\color{#3D99F6}{x}}^{\color{#3D99F6}{x}}= 2015, then which answer is equivalent to x x x x ? \Huge \color{#3D99F6}{x^{x^{x^x}}}?

x x 2015 x^{x^{2015}} 2015 x x { 2015 }^{x^x } 2015 2015 { 2015 }^{ 2015 } All the given choices are correct

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2 solutions

Sandeep Bhardwaj
Oct 9, 2015

Reminding the notion of tower rule , the powers are evaluated from the top (of the tower of powers) down to its bottom.

Substituting the value x x = 2015 { x }^{ { x }}=2015 into the required expression at the top, we get x x 2015 { x }^{ { x }^{{ 2015 } } } . \square

CAUTION:

We can't evaluate it further to 201 5 2015 2015^{2015} because we work from top to down, therefore taking the previous step, first we need to evaluate x 2015 x^{2015} , unknown to us. Hence we can't jump to the conclusion of 201 5 2015 2015^{2015} .

enjoy!

No tenía conocimiento de esta regla o propiedad de las potencias. Puede darme una referencia para leer acerca de esto. Gracias.

Jaime Maldonado - 5 years, 8 months ago

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En el link de Sandeep Bhardwaj tienes las propiedades, pero si quieres saber más sobre el tema, te recomiendo el siguiente artículo de Wikipedia: https://en.m.wikipedia.org/wiki/Tetration

Gon. E. B. - 5 years, 7 months ago

log(x^{x^{x^{x}}})=x^{x}log(x^{x}). As, x^{x}=2015, so, log(x^{x})=log(2015). So, log(x^{x^{x^{x}}})=2015 \times log(2015)=log(2015^{2015}). Hence, x^{x^{x^{x}}}= 2015^{2015}.

What is the error in the above?

Abhijit Pendse - 5 years, 7 months ago

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You cannot take the coefficient of the log of x^{x^{x^{x}}} as x^{x}, because it is not the power of the brackets of the log, the power of the log's brackets is x^{x^{x}} because there's only one x in the base. If it was {x^x}^{x^x} it would've worked as this will be simplified to x^{x{x^x}}.

Yahya Khashaba - 5 years, 2 months ago

You actually can evaluate it further. Using Excel's Goalseek function, you can find that for x^x to = 2015, x must equal 4.83072469611793. Then it's just a matter of plugging it in to get 1.28303*10^77. I'm sure there's some truncated digits, though I'm an engineer so I only deal with rounded numbers anyways :)

What I'm wondering, is there a function for finding x for a problem like this, or iteration the only way to find a solution?

Andrew Cullingham - 5 years, 2 months ago

Tengo dudas al respecto.

Jaime Maldonado - 5 years, 8 months ago

But isn't (((x^x)^x) ^x) = x^(x^3)?

Sidh Satam - 5 years, 8 months ago

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Try pluggin in some values eg 2 and see what happens

Chad Frederick - 5 years, 8 months ago

(((x^x)^x)^x) is but the question was (x^(x^(x^x))). You always work from top to bottom.

Davis Parks - 5 years, 2 months ago

Yes, but x x x x = x ( x ( x x ) ) x^{x^{x^x}} = x^{(x^{(x^x)})}

x x x x ( ( ( x x ) x ) x ) x^{x^{x^x}} \neq (((x^x)^x)^x)

You have to respect the operators priority.

Valentin Michelet - 5 years, 8 months ago

x x = 2015 x x x x = x x 2015 \huge \color{#D61F06}{x^{x} = 2015} \quad \Rightarrow x^{x^{{\color{#D61F06}{{x^x}}}}} = \boxed{x^{x^{{\color{#D61F06}{2015}}}}}

Note that by convention, calculations start from top to bottom.

can u explain this questio??

Nazir Balti - 5 years, 8 months ago

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The following article from Wikipedia talks about towers of powers: https://en.m.wikipedia.org/wiki/Tetration

Hope that helps...

Gon. E. B. - 5 years, 7 months ago

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For your this act of kindness I follow back you Thanks!!!

Nazir Balti - 5 years, 7 months ago

I have added a statement. Is that what you are expecting?

Chew-Seong Cheong - 5 years, 8 months ago

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But Where Is Your statement?? thanks!

Nazir Balti - 5 years, 7 months ago

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@Nazir Balti "Note that by convention, calculations start from top to bottom."

Chew-Seong Cheong - 5 years, 7 months ago

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