Mylee Underwood

2022-07-09

A question on inequality equivalence
I'm doing a textbook question and I'm stuck in the very last end...
I got the following function on S:
$f\left(S\right)=S-\frac{q\left(1-{q}^{S}\right)}{p}+{q}^{S}c$, where $p:=1-q$ and $0.
Now the first forward difference of f(S), so $\mathrm{\Delta }\left(S\right):=f\left(S+1\right)-f\left(S\right)$ is $\mathrm{\Delta }\left(S\right)=1-{q}^{S}\left(pc+q\right)$ and this function is increasing in S.
Now I need to find the value of S that satisfies $\mathrm{\Delta }\left(S\right)\ge 0$.
We can then write an inequality:
${q}^{S}\le \frac{1}{\left(pc+q\right)}$
But now when I want to look for S, we can take logarithms and find:
$S\le \frac{-ln\left(pc+q\right)}{ln\left(q\right)}$
But the textbook says that the equivalent inequality to ${q}^{S}\le \frac{1}{\left(pc+q\right)}$ is: $S\ge \frac{ln\left(pc+q\right)}{-ln\left(q\right)}$
Why is this?
They also say in the question: "using the fact of $\mathrm{ln}\left(q\right)<0$.
Any help is appreciated!

Kaylie Mcdonald

Expert

Step 1
The reason why the inequality sign is reversed is because $q<1$, so ${\mathrm{log}}_{q}\left(x\right)$ is actually a decreasing function.
Step 2
Therefore applying log to both sides will reverse the direction of inequality.

EnvivyEvoxys6

Expert

Explanation:
When you divide by a negative number the direction of the inequality get reversed. For example: $5<10$ but when dividing by -1 the inequality we have to change the direction: $-5>-10$.

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