Solutions — Exp/Log 13G Worksheet

Year 11 Methods Unit 1 · Cambridge §13G · Mr Wong

ANSWER KEY

Part A — Translations of $y=\log_2(x)$

Q1. Argument $>0$ gives the domain; argument $=0$ gives the asymptote; argument $=a^{-k}$ gives the $x$-intercept.

a dom $x>0$; asy $x=0$; $x$-int $(1,0)$

b dom $x>2$; asy $x=2$; $x$-int $(3,0)$ ($x-2=1$)

c dom $x>-5$; asy $x=-5$; $x$-int $(-4,0)$

d dom $x>0$; asy $x=0$; $x$-int $(\tfrac{1}{2},0)$ ($\log_2(x)=-1$)

Part B — Reflections

a $y=-\log_2(x)$: dom $x>0$, asy $x=0$, $x$-int $(1,0)$, decreasing

b $y=\log_2(-x)$: dom $x<0$, asy $x=0$, $x$-int $(-1,0)$, decreasing

c $y=\log_{1/2}(x)$: dom $x>0$, asy $x=0$, $x$-int $(1,0)$, decreasing (same graph as part a)

Q3. Change of base with base $2$: $\log_{1/2}(x)=\dfrac{\log_2(x)}{\log_2(1/2)}=\dfrac{\log_2(x)}{-1}=$ $-\log_2(x)$. ✓

Part C — Combined transformations

Q4. $y=\log_2(3x-2)$.

· dom: $x>\tfrac{2}{3}$; asy: $x=\tfrac{2}{3}$; $x$-int: $3x-2=1\Rightarrow$ $(1,0)$; second point $(\tfrac{4}{3},1)$ (where $3x-2=2$); range $\mathbb{R}$; increasing.

Q5. $y=\log_{1/2}(x+1)-2=-\log_2(x+1)-2$.

· dom: $x>-1$; asy: $x=-1$; $x$-int: $\log_2(x+1)=-2\Rightarrow x+1=\tfrac{1}{4}\Rightarrow$ $(-\tfrac{3}{4},0)$; $y$-int: $-\log_2(1)-2=$ $(0,-2)$; decreasing; range $\mathbb{R}$.

Part D — Inverse functions

Q6. $f(x)=2^x-5$. Range of $f$: $f(x)>-5$.

· $y=2^x-5 \Rightarrow x=2^y-5 \Rightarrow$ $f^{-1}(x)=\log_2(x+5)$, dom $(-5,\infty)$, range $\mathbb{R}$

Q7. $f(x)=2\cdot 5^x+2$. Range of $f$: $f(x)>2$.

· $x=2\cdot 5^y+2 \Rightarrow 5^y=\tfrac{x-2}{2} \Rightarrow$ $f^{-1}(x)=\log_5\!\left(\tfrac{x-2}{2}\right)$, dom $(2,\infty)$, range $\mathbb{R}$

Q8. $f(x)=\log_2(x+3)-2$, dom $(-3,\infty)$.

· $x+2=\log_2(y+3) \Rightarrow y+3=2^{x+2} \Rightarrow$ $f^{-1}(x)=2^{x+2}-3$, dom $\mathbb{R}$, range $(-3,\infty)$

Q9. $f(x)=\log_3(x)$ on $\mathbb{R}^+$.

· $x=\log_3(y) \Rightarrow y=3^x$. $f^{-1}(x)=3^x$, dom $\mathbb{R}$, range $\mathbb{R}^+$

Challenge

Q10. $y=2^x$ vs $y=x$. At $x=0$, $2^0=1>0$; at $x=1$, $2>1$; at $x=2$, $4>2$; etc. The gap grows because $2^x$ grows faster than $x$ for all $x>0$, and for $x<0$ we have $2^x>0>x$. So $2^x>x$ for every real $x$ — the exponential never touches $y=x$. Reflecting in $y=x$ flips "above" to "below", so $y=\log_2(x)$ lies entirely below $y=x$ for all $x>0$ — and the two function graphs never intersect each other.