Evaluate $\frac{x^2}{y}$ where $x=a^{a^{a}}$ and $y=a^{a^{2a}}$
1.$1$
2.$x^{a^{a}}$
3.$x^{1-a^a}$
4.$x^{2-a^a}$
My solution:
$x^2=a^{a^{a}}*a^{a^{a}}=a^{2a^{a}}$
$\frac{x^2}{y}=\frac{a^{2a^{a}}}{a^{a^{2a}}}=a^{2a^a-a^{2a}}=x^{2-a^a}$
I don't know where I am mistaked but our teacher gave the answer $1$. Could you please tell my mistake.
Your answer is correct. As written in a comment to a previously deleted answer, $(r^s)^t=r^{st}$. In this case, $r=a,s=a^a,$ and $t=2$. So $x^2=a^{2a^a}$, not $a^{a^{2a}}$. And using this rule again in reverse to add in a step you did not include in your solution
$$a^{2a^a-a^{2a}}=a^{a^a(2-a^a)}=(a^{a^a})^{2-a^a}=x^{2-a^a}$$