Raising a logarithmic function by e

Question

For example if we have $$ \ln{y}=a\ln{x}$$

If we raise both sides to the power of e:

$$ y = e^a .e^{\ln{x}} = e^ax$$

However by using log rules we get a different solution

i.e. by letting $$a\ln{x} = ln({x^{a}}) $$

then $$ y = x^a $$

Which of the two solutions is correct? Both seem to not violate anything but give different answers.

Answer 1

$$\ln{y}=a\ln{x} \implies y= e^{a\ln(x)}=x^a$$

You are confusing two rules

$$e^{a+b}=e^a.e^b$$ and $$e^{ab}=(e^a)^b=(e^b)^a$$

Answer 2

see that, $e^{a \ln(x)} \ne e^{a}.e^{lnx}$

Answer 3

Unless you believe that $t^{pq}=t^p\cdot t^q$, your argument is faulty.

This already fails for $p=q=1$, because the left-hand side is $t$ and the right-hand side is $t^2$. As soon as $t\ne1$, they differ.

The correct procedure is to note that $a\ln x=\ln(x^a)$, so you get $$ \ln y=\ln(x^a) $$ and therefore $y=x^a$.