3.2 Logarithmic and exponential functions

Syllabus
9709–2028–2029
Topic
3.2
Level
A2

Learning objectives

Translate index statements with the three logarithm laws

For $a>0$, $a e1$ and $y>0$:a^x=y\quad\Longleftrightarrow\quad \log_a y=x.Hence $\log_a1=0$ and $\log_a a=1$.

Structure Logarithm law (positive arguments)
product loga(MN)=logaM+logaN\log_a(MN)=\log_aM+\log_aN
quotient loga(M/N)=logaMlogaN\log_a(M/N)=\log_aM-\log_aN
power loga(Mp)=plogaM\log_a(M^p)=p\log_aM

Record the positivity conditions before combining or expanding logs. Move coefficients into powers when useful, combine to one logarithm, then translate back to index form.

For $x>0$,\log_2(8x)-\log_2x=\log_2(8)=3.

log(a+b)\log(a+b) does not split. Change-of-base formulae are explicitly excluded from this objective.

Read exponential and natural-log graphs as inverse shapes

y=e^x\quad\Longleftrightarrow\quad x=\ln y.Thus $\ln(e^x)=x$ for real $x$, while $e^{\ln x}=x$ requires $x>0$. Their graphs reflect in $y=x$.

Graph Domain Range Intercept/asymptote
y=ekxy=e^{kx}, $k
e0allreal| all realx|y>0|(0,1);horizontalasymptote; horizontal asymptotey=0$
y=lnxy=\ln x x>0x>0 all real yy (1,0)(1,0); vertical asymptote x=0x=0

For k>0k>0, ekxe^{kx} increases; for k<0k<0, it decreases. Both stay positive and approach, but never cross, the xx-axis in one direction.

$3e^{2x}=12$ gives $e^{2x}=4$, so $x= frac12\ln4$.

An exponential graph has no xx-intercept, and lnx\ln x is undefined for x0x\le0. Do not treat ln(x2)=2lnx\ln(x^2)=2\ln x as valid when x<0x<0.

Bring an unknown exponent down with logarithms

Rearrange until each exponential expression is positive, take ln\ln of both sides, use ln(ag(x))=g(x)lna\ln(a^{g(x)})=g(x)\ln a, then solve the resulting algebraic equation and check in the original.

5^{2x-1}=12\quad\Rightarrow\quad(2x-1)\ln5=\ln12\quad\Rightarrow\quad x= rac{1+\ln12/\ln5}{2}.

Base form Monotonic direction
au<ava^{u}<a^{v} with a>1a>1 u<vu<v
au<ava^{u}<a^{v} with 0<a<10<a<1 u>vu>v

If terms such as a2xa^{2x} and axa^x occur together, set t=axt=a^x with t>0t>0, solve the polynomial in tt, reject non-positive roots, then take logs to recover xx.

Logs can only be taken after both sides are positive. An inequality reverses for a decreasing base 0<a<10<a<1, not merely because logarithms were used.

Linearising a relationship makes a model testable with a straight-line graph

Transform variables so the model becomes Y=mX+c. For y=ab^x, plotting ln y against x gives gradient ln b and intercept ln a; for y=ax^n, plotting ln y against ln x gives gradient n.

Transform measured uncertainties and units consistently, then interpret the gradient and intercept in the original parameters. A straight plot supports the model but does not prove causation.

If ln y=0.7x+1.2, then a=e^{1.2} and b=e^{0.7} for y=ab^x.

The intercept is not always the original constant; it may be ln a or another transformed quantity.