Keep the denominator sign in view · source reference

Complete question

Solve $(2x-18)/(x+3)>x-6$ algebraically. [3] Hence solve $(\ln(x^2)-18)/(\ln x+3)>\ln x-6$. [3]

[2009/JJC/Prelim/I/6b]

Printed working

The first inequality is labelled (*) in the printed solution.

$$\begin{aligned} \frac{2x-18}{x+3}&>x-6\\ \frac{2x-18}{x+3}-(x-6)&>0\\ \frac{2x-18-(x+3)(x-6)}{x+3}&>0\\ \frac{-x^2+5x}{x+3}&>0\\ \frac{x(x-5)}{x+3}&<0\\ x(x-5)(x+3)&<0,\qquad x\ne-3. \end{aligned}$$

Thus $x<-3$ or $0<x<5$.

The printed note contrasts this with $x<-3$ and $0<x<5$, which would give $x\in\varnothing$.

For the logarithmic inequality,

$$\begin{aligned} \frac{\ln x^2-18}{\ln x+3}&>\ln x-6\\ \frac{2\ln x-18}{\ln x+3}&>\ln x-6. \end{aligned}$$

Using earlier result (*) with $x$ replaced by $\ln x$,

$$\ln x<-3\quad\text{or}\quad0<\ln x<5 \quad\Longrightarrow\quad0<x<e^{-3}\quad\text{or}\quad1<x<e^5.$$

Teacher margin notes

As $x$ is a variable, $x+3$ can be either negative or positive. As such, DO NOT cross-multiply.

Express the inequality in the form $f(x)/g(x)>0$, i.e. make one side of the inequality zero by bringing all terms from one side of the inequality to the other.

Multiply both sides of the inequality with $(x+3)^2$. Since $(x+3)^2$ is always positive for $x\ne-3$, the inequality sign remains unchanged after the multiplication.

Solution is written as “… or …”, NOT “and”: “or” is the union between the ranges, while “and” is the intersection of the ranges.

Source diagrams

Provenance

Question page references: 1. Solution page references: 1. The original compilation header and printed page numbers are preserved in the edition notes. Original PDFs and page images remain only in the private QA archive.