Basic doubt about exponents.

Question

I have a doubt regarding this:

$\begin{aligned} \ln y &=\ln \left(x^{x^{x}}\right) \\ &=x \ln \left(x^{x}\right) \\ &=x \cdot x \cdot \ln x \\ &=x^{2} \cdot \ln x \\ &=\ln \left(x^{x^{2}}\right) \neq \ln \left(x^{x^{x}}\right) \end{aligned}$

I KNOW that I did it wrong ( in step 3) but can any body explain me what exactly I did wrong and point out the exact fundamental understanding that I lack. Please try to be extremely specific while explaining the error.

If u want more info on the question, feel free to comment.

Answer 1

Your error was due to incorrect association of the exponents:

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

As you know,

$$ x \ln(u) = \ln\left(u^x\right). $$

Observe that

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

Also observe that

$$ \left(x^x\right)^x = x^\left(x^2\right). $$

And note that you are inconsistent in the way you associate exponentiation, because you treated $x^{x^x}$ as $\left(x^x\right)^x$ but you treated $x^{x^2}$ as $x^\left(x^2\right).$

Answer 2

The problem is that $$\log(x^{x^x})=x^x\log x\neq x^2\log x.$$ The logarithmic law you wanted to use is $$\log(a^b)=b\log a.$$ You set $a=x^x$ and $b=x$. The problem is that $a^b\neq x^{x^x}$, $a^b=x^{x^2},$ because $[(a^b)]^c=a^{bc}$. The correct way is to set $a=x$ and $b=x^x$.