Is this true? $e^{ab^{2}} = e^{a^{b^{2}}} = e^{2ab}$

Question

Is $$e^{ab^{2}} = e^{a^{b^{2}}} = e^{2ab}?$$

I'm only really curious to know if the first term equals the second term, I just wanted to show my steps.

Answer 1

No, it's not. In fact if this expression are equal then we would have: $$ab^2=2ab=a^{b^2}$$ but this is false for $a=3$ and $b=4$, so your hypotesis is false.

Answer 2

You need to be careful because exponentiation is not associative. In general, if you don't write parenthesis, the convention is that they associate to the left, i.e.$$ x^{y^z} = x^{(y^x)} \ne (x^y)^z $$ For your example:$$ \left(e^{ab}\right)^2 = \left(\left(e^a\right)^b\right)^2 = e^{2ab} $$ is true, but $$ e^{\left((ab)^2\right)} \ne e^{\left(a^{\left(b^2\right)}\right)} \ne e^{2ab} $$

Answer 3

Let $a=b=1$. ${}{}{}{}{}{}{}{}$

Answer 4

While $x+y+z$ can be grouped as either $(x+y)+z$ or $x+(y+z)$, we usually do not bother - because both refinements produce the same result.

However, with exponentiation, we have $$\tag1x^{(y^z)}\ne\bigl(x^y\bigr)^z $$ in general (e.g., $2^{(2^3)}=2^8=256$ and $\bigl(2^2\bigr)^3=4^3=64$. Being lazy, we want to be allowed to write one of the two expressions in $(1)$ without parentheses. As it always holds that $$\bigl(x^y\bigr)^z=x^{yz}, $$ it would be wasteful to let ${x^y}^z$ stand for the right hand side in $(1)$. Hence, we agree to interpret $x^{y^z}$ as the left-hand side expression.

It seems that you fell for this trap and switched interpretations of nested exponentiation in your computation.

Answer 5

You can say $\ e^{ab^2}$= $\ (e^a)^{b^2}, $ $\quad$ $\ e^{2ab}$=$\ ((e^{2})^a)^b$