Log Property Question

Question

How does $$e^{-\lambda k \frac{1}{\lambda}ln(2)} = e^{ln(2^{-k})} = \frac{1}{2^k}$$ by log properties?

Answer 1

Let's just look at the exponent (power) initially.

First the $\lambda$ and the $\frac 1 {\lambda}$ cancel out, that's just basic algebra. That gives $-k\ln(2)$.

A rule of logs is that $\ln(a^b) = b\ln a$, and it works in reverse too. The logarithm can be to any base too.

Applying that in reverse, the exponent becomes:

$$\ln 2^{-k}$$

Now, note that exponentiation is the inverse function of the logarithm, so taking $e$ to the power of the logarithm of something will just yield that "something", i.e. $e^{\ln y} = y$.

We therefore get $e^{\ln 2^{-k}} = 2^{-k}$.

Finally, recognise that $m^{-n} = \frac 1{m^n}$, which is a basic rule of exponents.

Then the final step is to simply recognise that $2^{-k} = \frac 1{2^k}$, as required.

Answer 2

We have $\text{log}(a^b) = b\text{log}(a)$. By using that we get $$e^{-\lambda k \frac{1}{\lambda}\text{log}(2)} = e^{-k\text{log}(2)} = e^{\text{log}(2^{-k})} = 2^{-k} = \frac{1}{2^k}.$$