Year 11 Methods — Exponentials & Logarithms (Unit summary)

Powers and roots

$a^{m/n}=\big(\sqrt[n]{a}\big)^{m}=\sqrt[n]{a^{m}}$

$a^{-n}=\dfrac{1}{a^{n}}$,   $a^{0}=1$

$a^{m}\cdot a^{n}=a^{m+n}$,   $\dfrac{a^{m}}{a^{n}}=a^{m-n}$,   $(a^{m})^{n}=a^{mn}$

When evaluating, do the root first — keeps the numbers small. e.g. $64^{2/3}=(\sqrt[3]{64})^{2}=4^{2}=16$.

Shape of $y=a^{x}+\text{transformations}$

$y=a^{x}\;(a>1)$: domain $\mathbb{R}$, range $\mathbb{R}^{+}$, $y$-int $(0,1)$, asymptote $y=0$, increasing

$y=a^{-x}=(1/a)^{x}$: same shape, reflected in the $y$-axis (decreasing)

$y=A\cdot a^{n(x-h)}+k$: asymptote $y=k$, $y$-int found by $x=0$

Vertical translations shift the asymptote. Use a table of values for the basic shape, then apply transformations point-by-point.

Choose your method

Same base: rewrite both sides as $a^{(\,\cdot\,)}$, equate indices.

Quadratic in $a^{x}$: let $u=a^{x}$, solve the quadratic, reject any $u\le 0$ (since $a^{x}>0$).

Inequalities: $a^{x}>k$ — solve like an equation; flip the inequality if the base is $0<a<1$.

Different bases that can't be matched? → use logs (§13F).

Definition + the seven laws

$\log_{a}(x)=y\iff a^{y}=x$.   Domain $x>0$.

$\log_{a}(mn)=\log_{a}(m)+\log_{a}(n)$

$\log_{a}\!\left(\tfrac{m}{n}\right)=\log_{a}(m)-\log_{a}(n)$

$\log_{a}(m^{p})=p\log_{a}(m)$,   $\log_{a}(a)=1$,   $\log_{a}(1)=0$

Change of base: $\log_{a}(x)=\dfrac{\log_{b}(x)}{\log_{b}(a)}$

When bases don't match

Take a log of both sides (any base — usually $\log_{10}$ on CAS or $\log_{a}$ to match a base in the question).

e.g. $5^{x}=12 \Rightarrow x=\log_{5}(12)\approx 1.54$

Always check domain on log questions — anything that makes the argument $\le 0$ is rejected.

For $a^{x}=b$ with $a>1$: exact $x=\log_{a}(b)$, decimal $x=\log(b)/\log(a)$.

Reflection in $y=x$

$y=\log_{a}(x)$ & $y=a^{x}$ are inverses → reflections in $y=x$.

$y=\log_{a}(x)$: domain $x>0$, range $\mathbb{R}$, $x$-int $(1,0)$, asymptote $x=0$ (the $y$-axis).

$y=\log_{a}(x-h)+k$: asymptote $x=h$, $x$-int where $x-h=a^{-k}$.

$\log_{1/a}(x)=-\log_{a}(x)$ — reflection in the $x$-axis.

To find an inverse: write $y=f(x)$, swap $x$ and $y$, solve for $y$, then state the domain (= range of $f$).