An exponential reflected at its zero · source reference

Complete question

(i) Sketch $y=|(1/3)^{2x}-18|$, finding exact intercepts. [3] (ii) Without a calculator, solve $|(1/3)^{2x}-18|\le9$ with full working, giving the exact interval or intervals in their simplest form. [3]

[2022/ACJC/Prelim/I/2]

Printed working

Part (i)

The graph of $y=|3^{-2x}-18|$ has intercepts $(0,17)$ and $(-\ln18/(2\ln3),0)$, and horizontal asymptote $y=18$.

$$3^{-2x}-18=0\quad\Longrightarrow\quad2x=\frac{\ln18}{\ln(1/3)} \quad\Longrightarrow\quad x=\frac{\ln18}{-2\ln3}.$$

Part (ii): Method 1 — graphical

The source shows the same curve together with $y=9$; points A and B mark its two intersections with that line.

At A: $3^{-2x}-18=9\Rightarrow3^{-2x}=3^3\Rightarrow x=-3/2$.

At B: $-[3^{-2x}-18]=9\Rightarrow3^{-2x}=3^2\Rightarrow x=-1$.

Therefore $-3/2\le x\le-1$.

Part (ii): Method 2 — algebraic

$$\begin{aligned} |3^{-2x}-18|&\le9\\ -9\le3^{-2x}-18&\le9\\ 9\le3^{-2x}&\le27\\ 3^2\le3^{-2x}&\le3^3\\ 2\le-2x&\le3\\ -\frac32\le x&\le-1. \end{aligned}$$

Alternatively,

$$\left(\frac13\right)^{-2}\le\left(\frac13\right)^{2x}\le\left(\frac13\right)^{-3} \quad\Rightarrow\quad-2\ge2x\ge-3\quad\Rightarrow\quad-\frac32\le x\le-1.$$

Source diagrams

Provenance

Question page references: 3. Solution page references: 16. 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.