Solve the inequality with logarithm

fractionsinequalitylogarithms

The inequality is: $$\frac{1}{\log_{(x-1)}\frac{x}{20}}\ge-1$$

I made a plot of the function $f(x)=\frac{1}{\log_{(x-1)}\frac{x}{20}}+1$ and it looks the answer is $x\in(1,5]\cup(20,+\infty)$. Using logarithm definition I conclude that the argument $\frac{x}{20}>0$, the base $x-1>0$ and $x-1\ne1$. Also I made a couple of transformations of initial inequality to retrieve definite answer but it doesn't quite match what I see on the plot.

I am looking for a specific solution steps required to get the answer.

Best Answer

As hinted by KM101 and myself: $$ \log_ab = \frac{\ln b}{\ln a} $$ Then your inequality is equivalent to: $$ \frac{\ln (x-1)}{\ln\frac{x}{20}} \geq -1 $$

Note that we must have $x\neq20$ and $x>1$ for all those quantities to exist.

Now, there are two cases to consider:

  • $\ln\frac{x}{20} > 0$ ie. $\frac{x}{20}>1$ and then: $$ \ln(x-1) \geq \ln\frac{20}{x} $$ (we already set $x>1$ so $x\neq 0$)

and by applying $x\mapsto e^x$ (which is increasing on $\mathbb{R}$) on each side: $$ x-1\geq \frac{20}{x} $$

No need to do much more here since we have $\frac{x}{20}>1$ ie. $x>20$, the former inequallity is always verified ($x-1>19$ and $\frac{20}{x}<1$). Thus, the inequality holds for all $x>20$.

  • $\ln\frac{x}{20} < 0$ ie. $\frac{x}{20}<1$ and then: $$ \ln(x-1) \leq \ln\frac{20}{x} $$

and by applying $x\mapsto e^x$ (which is increasing on $\mathbb{R}$) on each side: $$ x-1\leq \frac{20}{x} $$ which yields $$ x^2 - x -20 \leq 0 $$ $-4$ and $5$ are obvious roots (if not use standard methods to find them). Then the sign of the polynomial is easy to find: it is positive or zero for $x\leq-4$ and $x\geq5$ and negative or zero for $-4\leq x\leq5$. What matters for us given the inequality is the latter case. But from former hypotheses, we must also have $x>1$ and $x<20$. Thus, the inequality holds for $1<x\leq 5$.

In the end, the two distinct cases indeed yield $x\in (1,5]\cup(20,\infty)$.

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