1
$\begingroup$

I have two functions 1) $y=x^{x^{x}}$ 2) $y=(x^x)^x$

These two functions seem same to me and I just see it as a mere difference of writing style but when I graph it using an online graph plotter they have different curves also when I find their derivatives using logarithmic differentiation I get different results.For 1 and 2 I got $dy/dx$ as $x^{x^{x}}[x^x\cdot\ln(x)[1+\ln(x)]+x^{(x-1)}]$ and $(x^x)^x[x[2\ln(x)+1]]$ respectively

So,my question is ,Are these two functions really different,if yes ,how?If no,how can you justify their similar looking expressions?

$\endgroup$
2
  • 1
    $\begingroup$ They are indeed different. The second one is $x^{x^2}$. $\endgroup$
    – user228113
    Feb 13, 2017 at 8:08
  • $\begingroup$ @G.Sassatelli are x^x^x and (x^x)^x ,different for you? ,2nd function shown in question has Latex as (x^x)^x.Now what is your say? $\endgroup$ Feb 13, 2017 at 8:12

3 Answers 3

1
$\begingroup$

They are different, because $$3^{3^3} = 3^{(3^3)} = 3^{27} = 7\,625\,597\,484\,987$$ whilst $${(3^3)}^3 = 27^3 = 19\,683 = 3^9 = 3^{(3^2)}$$

One of general properties of exponentiation is $$(a^b)^c = a^{(b\cdot c)}$$ which corresponds to $$\log (a^b)^c = c\cdot\log a^b = c\cdot(b\cdot\log a) = (c\cdot b)\log a = \log a^{b\cdot c}$$ hence $$(x^x)^x = x^{(x\cdot x)} = x^{(x^2)} \ne x^{(x^x)}$$

$\endgroup$
3
  • $\begingroup$ I understand these facts sir but the second function has latex (x^x)^x.Still you will say it is equivalent to (x)^{x^2} ? $\endgroup$ Feb 13, 2017 at 8:19
  • $\begingroup$ Sure it is. Please see my expanded answer. $\endgroup$
    – CiaPan
    Feb 13, 2017 at 8:30
  • $\begingroup$ but if I wanted to set (x^x) to the powered x .Then how will you express it? or does it mean x^(x^2) is same as saying (x^x) powered x? $\endgroup$ Feb 13, 2017 at 8:31
1
$\begingroup$

Note that the exponent of $y=x^{x^x}$ can be represented as $k=x^x$ so $y=x^k$. For $y=(x^x)^x$ the exponent inside of the bracket is multiplied with the outside exponent so $k=x\cdot x= x^2$ then $y=x^k=x^{x^2}$

$\endgroup$
4
  • $\begingroup$ I understand these facts sir but the second function has latex (x^x)^x.Still you will say it is equivalent to (x)^{x^2} ? $\endgroup$ Feb 13, 2017 at 8:17
  • $\begingroup$ Both exponent are multiplied (exponent's rule) so $(x^x)^x = x^{x\cdot x} = x^{x^2}$. $\endgroup$
    – kub0x
    Feb 13, 2017 at 8:20
  • $\begingroup$ but if I wanted to set (x^x) to the powered x .Then how will you express it? $\endgroup$ Feb 13, 2017 at 8:25
  • 1
    $\begingroup$ $x^x$ to the $x-th$ power is $(x^x)^x$, as said before this yields $x^{x\cdot x}=x^{x^2}$. You could express $k=x^x$ then $k^x$ is what you are looking for. $\endgroup$
    – kub0x
    Feb 13, 2017 at 8:33
1
$\begingroup$

$x^{x^x}$ is normally parsed as $x^{(x^x)}$. Which is different than $(x^x)^x$ which is $x^{x^2}$.

To see the difference try $3^{3^3}=3^{27}$ whereas $(3^3)^3=27^3$. The first is on the order of $10^{12}$ where as the second is on the order of $10^5$.

$\endgroup$
2
  • $\begingroup$ I understand these facts sir but the second function has latex (x^x)^x.Still you will say it is equivalent to (x)^{x^2} ? $\endgroup$ Feb 13, 2017 at 8:19
  • 1
    $\begingroup$ @KartikWatwani Yes. This is easiest seen on an example. Note that $(4^4)^4 = 4^4 * 4^4 * 4^4 * 4^4=4^{4+4+4+4}=4^{4*4}=4^{4^2}$ in this holds in general if you take $(n^n)^n$ that's the same $n^n\cdots n^n$ $n$ times which then is $n^{n+\cdots n}$ where the addition is $n$ times and so is $n^{n*n}=n^{n^2}$. This is obviously not even close to a proof but it gives you an inkling of the idea behind the definition. $\endgroup$
    – DRF
    Feb 13, 2017 at 8:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.