Why is $\sqrt x\ne\frac1{x^2}$?

Question

Which step is illegal?

1. $\sqrt x=x^{\frac12}$

2. $\frac12=2^{-1}$

3. $x^{\frac12}= x^{2^{-1}}$

4. $a^{b^c}=a^{bc}$

5. $x^{2^{-1}}=x^{-2}$

6. $a^{-b}=\frac1{a^b}$

7. $x^{-2}=\frac1{x^2}$

Am I correct in assuming that it is step 3, because $a^{b^c}$ is not the same as $a^d$ where $d=b^c$ but rather $a^d=a^{(b^c)}$?

Answer 1

For sake of natation:

$(a^b)^c=a^{bc} $

But $a^{(b^c)}\ne a^{bc} $

So $x^{\frac 12}=x^{(2^{-1})}\ne (x^2)^{-1}=\frac 1 {x^2} $.

Others explained but I hope notation makes it clear.

Answer 2

The illegal step is Step 4. Exponentiation is not associative, as is suggested in one of the comments you got, i.e.

$$a^{bc} = ({a^b})^c \neq a^{b^c}$$

By the way, I don't know what you would like to show with steps 6 and 7. Given that your step 4 is wrong anyway, the correctness of latter steps does not really matter...

Answer 3

In step 3, you have $x^{(2^{-1})}$, but when you apply step 4 to step 3 to get step 5, you treat it as $(x^2)^{-1}$, because step 4 really is $(a^b)^c = a^{bc}$