Something Fishy Here

Algebra Level 1

True or false:

$\huge \displaystyle \sqrt[3]{3^{3^3}} = 3^3 .$

True False

3 solutions

Nihar Mahajan
Apr 24, 2016

Relevant wiki: Rules of Exponents - Algebraic

$\huge \begin{aligned} \sqrt[x]{x^{x^x}} = \left(x^{x^x}\right)^{\frac{1}{x}}= \left(x^{x^x}\right)^{x^{-1}}= x^{x^x\times x^{-1}} &= x^{x^{x-1}} \\ &\neq x^x \end{aligned}$

If you have any queries, first review Rules of Exponents wiki and then comment below.

Moderator note:

Great question!

Again fish!!! In your title :p :p

lol... XD

Atul Shivam - 5 years, 1 month ago

What do you mean by "Again fish"?

Nihar Mahajan - 5 years, 1 month ago

Again your favorite thing in the title XD!!!

Atul Shivam - 5 years, 1 month ago

@Atul Shivam – Who said fish is my favorite thing? :P

Nihar Mahajan - 5 years, 1 month ago

@Nihar Mahajan – I know everything!!! :P

Atul Shivam - 5 years, 1 month ago

but this relation is true for 2 so you can't categorize it under false for all positive integers

Osama Rashid - 5 years, 1 month ago

The question stated "All positive real values of $x$ ."

It is not true for all of them, so the answer is $\boxed{\text{False}}$

Hung Woei Neoh - 5 years, 1 month ago

Just the main thing is, $\sqrt[x]{{x}^{x^x}} = (x^{x^x})^{\frac{1}{x}}$ Rest follows!!

akash patalwanshi - 5 years, 1 month ago

To me, it seems as though it can be true or false depending on which positive real number you choose. The statement is true for cases where x is equal to 1 or 2, and false for cases where x is neither 1 nor 2. To claim that an open statement like that is always true or always false is illogical. If it said something like, "for all x in the set of positive reals, P(x)", then I could see an argument for claiming that the existence of counterexamples would cause that closed statement to be false. But as we are not told any restrictions on x other than the fact that it is a positive real number, there is insufficient data to produce a definite answer.

Shawn Franchi - 5 years, 1 month ago

It will be correct for 2 so it can't be characterised as false

Ayush Kumar - 3 years, 6 months ago

Correct :+1: upvoted!!!

Atul Shivam - 5 years, 1 month ago

You are wrong! This is absolutely true. Your solution's 4th step doest really equals to 5th step, when multiplied by inverse function it becomes 1. So This is absolutely true. Try with any possible integer. :) Good Day!

Soham Ganguly - 5 years, 1 month ago

No nihar is absolutely correct!

Atul Shivam - 5 years, 1 month ago

Not at all. :)

Soham Ganguly - 5 years, 1 month ago

@Soham Ganguly – I dont find any mistake :(

Ashish Menon - 5 years, 1 month ago

No Brother. :) Try with examples or I will simplify it and give it to you :)

Soham Ganguly - 5 years, 1 month ago

@Soham Ganguly – Try with 0.1 and look brother

Ashish Menon - 5 years, 1 month ago

@Ashish Menon – Look brother. Mathematical Inequations can't be proved with examples. I am simplifying this for you.

x^x^x is equals to x^x-square. when rooting with power x this whole equation becomes x^x-square^1/x=x^x^(2-1)=x^x. I dont know how to format this :(

Soham Ganguly - 5 years, 1 month ago

@Soham Ganguly – So this might be the place where you are going wrong. ${\left({x^{x^x}}\right)}^{\frac{1}{x}} \neq x^x$
It is actually equal to $x^{x^x × \tfrac{1}{x}} = x^{x^x × x^{-1}} = x^{x^{x - 1}}$ We have to multiply the top exponent with all of the base given because Nihar's question's LHS can also be written as ${\left({\left(x\right)}^{x^x}\right)}^{\frac{1}{x}}$

Ashish Menon - 5 years, 1 month ago

@Ashish Menon – Just tell me how x^x^x^x^-1 is is equals to x^x^x^-1

Soham Ganguly - 5 years, 1 month ago

@Soham Ganguly – There is a multiplication sign between the last 2 x's of the LHS

Ashish Menon - 5 years, 1 month ago

@Ashish Menon – Now I am giving you a proper thing Wait! $x^{x^{x^{x^{-1}}}}$ = $x^{x^{1}}$ or $x^{x^{2} \cdot x^{-1}}$ = $x^{x^{2-1}}$ = $x^{x^{1}}$ = $x^{x}$

Soham Ganguly - 5 years, 1 month ago

@Soham Ganguly – Your first step is wrong

Ashish Menon - 5 years, 1 month ago

@Ashish Menon – When extreme top orders are multiplying what is the solution? How can x X x^-1 = x^-1

Soham Ganguly - 5 years, 1 month ago

@Soham Ganguly – That is the mistake! We have to multuply the entire exponent if the base with the one outside the bracket.

Ashish Menon - 5 years, 1 month ago

@Soham Ganguly – How $x \cdot x^{-1}$ is equals to $x^{-1}$ Oh I Did It :D ;)

Soham Ganguly - 5 years, 1 month ago

@Soham Ganguly – I didnt understand did you understand how to solve this problem?

Ashish Menon - 5 years, 1 month ago

@Soham Ganguly – Can I know why $x^{x^x} = x^{x^2}$ ?

Note that $x^{x^x} \neq (x^x)^x$

Hung Woei Neoh - 5 years, 1 month ago

@Hung Woei Neoh – Yes, maybe he has got tower rule confused.

Ashish Menon - 5 years, 1 month ago

@Ashish Menon – No, he definitely got his tower rules confused. It was rather obvious from his statement.

Do check out the rules to evaluating tower exponents. We always start from the highest exponent, that is:

$\Large x^{x^x} = x^{\left(x^x\right)} \neq (x^x)^{x}$

Ask yourself, is $\Large3^{3^3}$ and $\Large(3^3)^3$ the same thing?

Hung Woei Neoh - 5 years, 1 month ago

I know I am wrong :( But why would 16 people upvote a wrong solution and the challenge master note does not mention any mistake? :(

Nihar Mahajan - 5 years, 1 month ago

If this is not a sarcastic reply then I am really interested to report it to a mod. :) Its better if you go & do it. I saw this was shared by brilliant.org through fb page. This just ruins their reputations. :)

Soham Ganguly - 5 years, 1 month ago

@Soham Ganguly – I think you dont know but nihar himself is a mod

Ashish Menon - 5 years, 1 month ago

Try yourself by putting in any integer. It is NOT true.

Nicolai Kofoed - 5 years, 1 month ago

Lets take 2. 2^2^2=16. Square Root of 16 is equals to 4 i.e. 2^2 In Case of 3. 3^3^3=19683. Cube Root of 19683 is indeed 27 i.e 3^3

Soham Ganguly - 5 years, 1 month ago

@Soham Ganguly – Positive real number does not only mean that you have to use positive integers, substitute 0.1 and look you wont get the same answer. :)

Ashish Menon - 5 years, 1 month ago

@Ashish Menon – Yeah I know that. :)

Soham Ganguly - 5 years, 1 month ago

May I know how did you conclude that the 4rth step of Nihar's solution is not equal to his 5th step? I dont find any errors. Plz explain your viewpoint.

Ashish Menon - 5 years, 1 month ago

When multiplied by inverse it becomes 1. Sin X Sin^-1 is equals to 1. x X x^-1 is equals to 1. :)

Soham Ganguly - 5 years, 1 month ago

You got your rules of exponents confused. Here, I'll show you an example:

Say $x = 3$

$\Large\sqrt[x]{x^{x^x}}\\ =\Large\sqrt[3]{3^{3^3}}\\ =\Large\left(3^{27}\right)^{\frac{1}{3}}\\ =\Large3^{27 \times \frac{1}{3}}\\ =\Large3^9\\ \neq \Large3^3$

Hung Woei Neoh - 5 years, 1 month ago

Haha, nice

Ashish Menon - 5 years, 1 month ago

Joshua Lowrance
Mar 9, 2018

Shouldn’t it be 3^3^3 = (3^3)^3=27^3 rather than 3^27?

Doron Ophir - 2 years, 2 months ago

No. When dealing with multiple exponents (as it is in the case of $3^{3^3}$ ), you start from the highest exponent and work your way down. So, in the case of $3^{3^3}$ , it would be $3^{(3^3)}=3^{27}$ . You can also try it on a calculator, like desmos.com. Type in $3^{3^3}$ (without any parenthesis) and it should equal $3^{27}=7.625597485 \times 10^{12}$ .

Joshua Lowrance - 2 years, 2 months ago

Gia Hoàng Phạm
Nov 5, 2018

$\sqrt[x]{x^{x^x}}=(x^{x^x})^{\frac{1}{x}}=(x^{(x^x)})^{x^{-1}}=x^{(x^x)(x^{-1})}=x^{x^{x+-1}}=\boxed{\large{x^{x^{x-1}}}}$

Fixed typo in @Nihar Mahajan

Something Fishy Here

3 solutions

Moderator note:

lol... XD

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