Transition from exponential form to logarithmic form in an inequality This is my first question on math stack exchange and I am happy to join the community!
I would like to ask if someone could explain me in detail, how this transition is made (because I could not find any reference in the book):
$$\left(1-\frac{2}{n}\right)^k\le\frac{1}{3}$$ to $$k\ge\frac{\ln\frac{1}{3}}{\ln\left(1-\frac{2}{n}\right)}$$
Thank you very much in advance.
 A: $\ln\left(1-\frac{2}{n}\right)<0$ for all $n>3$ and $\ln$ is an increasing function.
Thus, we get
$$k\ln\left(1-\frac{2}{n}\right)\leq\ln\frac{1}{3}$$ or
$$k\geq\frac{\ln\frac{1}{3}}{\ln\left(1-\frac{2}{n}\right)}$$
A: Consider the equation
$$a^k\le b\tag{1}$$
where I'll assume $a$, $b>0$. The natural logarithm of $a^k$ is $k\ln a$.
For positive arguments, the logarithm preserves order, so $(1)$ is
equivalent to
$$k\ln a\le \ln b.\tag2$$
In our case $a=1-2/n$ and $b=1/3$, so $\ln a<0$ and $\ln b<0$.
So when we divide $(2)$ by $\ln a$, we must reverse the direction of inequality to get
$$k\ge\frac{\ln b}{\ln a}.$$
A: It may be more complicated, as there are counter-examples to your statement, such as 


*

*$n=1,k=3$  in which case $\left(1-\frac{2}{n}\right)^k = -1 \lt \frac13$ 

*$n=2, k=2$ in which case $\left(1-\frac{2}{n}\right)^k = 0 \lt \frac13$ 

*$n=-1, k=-2$ in which case $\left(1-\frac{2}{n}\right)^k = \frac19 \lt \frac13$ 


where $k\ge\dfrac{\log(\frac{1}{3})}{\log(1-\frac{2}{n})}$ is either meaningless or wrong.
Instead try something like


*

*If $n = 2$ then $\left(1-\frac{2}{n}\right) = 0 $ and $0^k \le \frac13$ is true for all $k > 0$

*If $n= 0$ then $\frac{2}{n}$ is not a real number
Otherwise take logarithms to get 
$$k\log\left(1-\frac{2}{n}\right)\le\log\left(\frac{1}{3}\right)$$


*

*If $n \lt 0$ then $\left(1-\frac{2}{n}\right) > 1 $ so $\log\left(1-\frac{2}{n}\right) > 0$ so dividing both sides by $\log\left(1-\frac{2}{n}\right)$ gives $k \le \dfrac{\log\left(\frac{1}{3}\right)}{\log\left(1-\frac{2}{n}\right)}$

*If  $n\gt 2$, and $0 \lt 1-\frac{2}{n} \lt 1$ so $\log\left(1-\frac{2}{n}\right)$ is  negative so dividing both sides by $\log\left(1-\frac{2}{n}\right)$ gives 
$k \ge \dfrac{\log\left(\frac{1}{3}\right)}{\log\left(1-\frac{2}{n}\right)}$
That leaves the case where $0 < n <2$  so $\left(1-\frac{2}{n}\right) < 0 $ and you have issues with powers of a negative number. 


*

*If $k$ is an odd integer and $0 < n <2$ then $\left(1-\frac{2}{n}\right)^k$ is negative so always less than $\frac13$ 


There might be also be a solution with $k$ an even integer (let's say $2m$) in which case you are also trying to solve $\left(\frac{2}{n} -1\right)^k \le \frac13$.  It will give similar solutions, restricted to $k$ being an even integer:


*

*if $n=1$ then $(-1)^{2m} =1$ so never less than or equal to $\frac13$ 

*if $1 < n<2$  then $k \le \dfrac{\log\left(\frac{1}{3}\right)}{\log\left(\frac{2}{n}-1\right)}$

*if $0 < n<1$  then $k \gt \dfrac{\log\left(\frac{1}{3}\right)}{\log\left(\frac{2}{n}-1\right)}$


Incidentally, $\log\left(\frac{1}{3}\right) = -\log\left(3\right)$, which might simplify the notation
A: taking the logarithm of both sides we get
$$k\ln\left(1-\frac{2}{n}\right)\le \ln\left(\frac{1}{3}\right)$$ from here
$$k\geq \frac{\ln\left(\frac{1}{3}\right)}{\ln\left(1-\frac{2}{n}\right)}$$
since $$0<\ln\left(1-\frac{2}{n}\right)<1$$
