Evaluate $\frac{x^2}{y}$ where $x=a^{a^{a}}$ and $y=a^{a^{2a}}$

Question

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.

I think your mistake came from the fact that $(a^{a^{a}})^2 = a^{a^{2a}}$. — John_dydx, Sep 06 '16 at 06:13

score 3 · Accepted Answer · answered Sep 06 '16 at 06:47

3

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

$$a^{2a^a-a^{2a}}=a^{a^a(2-a^a)}=(a^{a^a})^{2-a^a}=x^{2-a^a}$$

answered Sep 06 '16 at 06:47

Mike

13,318

Are you sure?Because I asked from my classmate and he gave the answer 1 too. – Taha Akbari Sep 06 '16 at 06:50
@TahaAkbari Assuming $a^{a^a}$ means $a^{(a^a)}$ and not $(a^a)^a$, yes. – Mike Sep 06 '16 at 06:55
@TahaAkbari The main problem is to understand who are the basis and the exponent. In my opinion Mike gives you the right answer, because if the basis was $a^a$ the brackets are needed. – InsideOut Sep 06 '16 at 07:01
@Mike of course. – Taha Akbari Sep 06 '16 at 07:06

Evaluate $\frac{x^2}{y}$ where $x=a^{a^{a}}$ and $y=a^{a^{2a}}$

1 Answers1