Why does this log simplify to this?

Question

I was answering a question on integrating a log, but I don't understand one part:

Why does $e^{1-ln4}$ simplify to $e^1/e^{ln4}$?

Answer 1

Assuming that $a$ is a positive real number you have that $a^{b-c} = a^b\times a^{-c}=a^b\times \dfrac{1}{a^c} = \dfrac{a^b}{a^c}$

This is simply an application of that using $e, 1, \ln 4$ in the place of $a,b,c$ respectively.

Be extremely cautious if the base of the exponent is not a positive real number. Things very frequently break.

Answer 2

It is$$ e^{1-\ln(4)}=e^1\cdot e^{-\ln(4)}=\frac{e^1}{e^{\ln(4)}}$$

Answer 3

I believe this is just using exponent rules maybe review them a little bit. But for your question $\frac{b^x}{b^y}$ $=$ $b^{x-y}$. Therefore, $\frac{e^1}{e^{ln4}}$ $=$ $e^{1-ln4}$.

Answer 4

This is the division rule for powers of the same base. If $b$,$x$ and $y$ are positive numbers, then $$\frac{b^x}{b^y} = b^{x-y}$$

So in your case, $$\frac{e^1}{e^{\ln4}} = e^{1-\ln4} $$