Removing expressions from exponentials

Question

I have this expression: $$\exp(E_f/kt) = \exp(Ev_1/kt) / \exp(Ev_2/kt)$$ In the equation after this, all the $kt$ terms are removed and we are left with: $$\exp(E_f) = \exp(Ev_1) / \exp(Ev_2)$$ How is that possible? What property did he use?

Answer 1

Let $E_f = a, kt = b, Ev_1 = c, Ev_2 = d$

So, $$e^{a/b} = \frac{e^{c/b}}{e^{d/b}}$$

$$(e^a)^{1/b} = \bigg(\frac{e^c}{e^d}\bigg)^{1/b}$$

[As, $x^{yz} = (x^y)^z$, $z = 1/b$]

$$(e^a) = \frac{e^c}{e^d}$$

So, $$\exp(E_f) = \frac{\exp(Ev_1)}{\exp(Ev_2)}$$

Answer 2

Hint: You can write $$\frac{e^{\frac{Ev_1}{kt}}}{e^{\frac{Ev_2}{kt}}}=e^{\frac{Ev_1}{kt}-\frac{Ev_2}{kt}}=e^{\frac{E}{kt}(v_1-v_2)}$$ so we get $$e^{\frac{E_f}{kt}}=e^{\frac{E}{kt}(v_1-v_2)}$$ and we obtain $$\frac{E_f}{kt}=\frac{E}{kt}(v_1-v_2)$$ thus $$E_f=E(v_1-v_2)$$

Answer 3

\begin{equation} e^{\frac{E_f}{kt}}=\frac{e^{\frac{Ev_1}{kt}}}{e^{\frac{Ev_2}{kt}}}=e^{\frac{Ev_1-Ev_2}{kt}} \end{equation} Then, it can be written \begin{equation} e^{E_f}=e^{Ev_1-Ev_2} \end{equation}

Answer 4

Just raise both sides to the $(kt)^\text{th}$ power. It’s that simple!