P3.3 - Exponentials and logarithms

Learning objectives

P3.3.1function ex and its graphThe function ex and its graph.; To include the graph of y = eax + b + c.P3.3.2function ln x and its graphThe function ln x and its graph; ln x Solution of equations of the form eax + b = p and as the inverse function of ex. ln (ax + b) = q is expected.P3.3.3Use logarithmic graphs to estimateUse logarithmic graphs to estimate Plot log y against log x and obtain a straight line where the parameters in relationships of the intercept is log a and the gradient is n. form y = axn Plot log y against x and obtain a straight line where the and y = kbx. intercept is log k and the gradient is log b.

The graph of $y=e^x$ is always positive, passes through $(0,1)$ and approaches the horizontal asymptote $y=0$ as $x\to-\infty$ . Its domain is $\mathbb R$ and its range is $(0,\infty)$ .

For $y=e^{ax+b}+c$ , with $a\ne0$ :

Feature	Result
horizontal asymptote	$y=c$
domain and range	$x\in\mathbb R$ ; $y>c$
$y$ -intercept	$(0,e^b+c)$
direction	increasing if $a>0$ ; decreasing if $a<0$

Writing $ax+b=a(x+b/a)$ shows that the inside change moves and horizontally scales the base graph; adding $c$ moves every output vertically and therefore moves the asymptote from $y=0$ to $y=c$ .

For $y=e^{-2x+1}+3,$ the asymptote is $y=3$ , the range is $y>3$ , and the curve decreases because the coefficient of $x$ is negative. At $x=0$ , $y=e+3$ , so $(0,e+3)$ fixes its vertical position.

The curve approaches its asymptote but never reaches it because $e^{ax+b}>0$ . Do not set the asymptote to $y=0$ after an outside shift, and do not use $b$ alone as a horizontal translation without first factoring $a$ .

The natural logarithm is the inverse of the exponential function: $\ln(e^x)=x$ and $e^{\ln x}=x$ for $x>0$ . Therefore the graphs of $y=e^x$ and $y=\ln x$ are reflections in $y=x$ .

Feature of $y=\ln x$	Value
domain	$x>0$
range	$\mathbb R$
intercept	$(1,0)$
vertical asymptote	$x=0$

$e^{ax+b}=p\Rightarrow x=\frac{\ln p-b}{a}\quad(p>0),\qquad \ln(ax+b)=q\Rightarrow x=\frac{e^q-b}{a}\quad(ax+b>0),\qquad a\ne0$

For $e^{3x-1}=7$ , take natural logs to get $3x-1=\ln7$ , so $x=(1+\ln7)/3$ . For $\ln(5-2x)=1$ , exponentiate to obtain $5-2x=e$ , hence $x=(5-e)/2$ ; this satisfies the required domain $5-2x>0$ .

A logarithm accepts only a positive argument. Do not write $\ln(u+v)=\ln u+\ln v$ , and do not take $\ln p$ when an equation has $p\le0$ ; in that case the real exponential equation has no solution.

A logarithmic plot turns a nonlinear relationship into a straight line. The chosen horizontal axis determines which model is being tested, while the gradient and intercept recover the original parameters.

Original model	Straight-line form	Plot	Gradient	Intercept
$y=ax^n$	$\log y=n\log x+\log a$	$\log y$ against $\log x$	$n$	$\log a$
$y=kb^x$	$\log y=x\log b+\log k$	$\log y$ against $x$	$\log b$	$\log k$

Using base-10 logarithms, an intercept $c$ means $a=10^c$ or $k=10^c$ . For the exponential plot, a gradient $m$ means $b=10^m$ . With natural logarithms, use $e^c$ and $e^m$ instead; keep one base throughout.

Suppose a graph of $\log_{10}y$ against $\log_{10}x$ has gradient $-\frac12$ and intercept $2$ . Then $\log_{10}y=-\frac12\log_{10}x+2,$ so $y=10^2x^{-1/2}=100x^{-1/2}.$

Estimate the gradient from two well-separated points on the straight line of best fit, not automatically from two raw data points. Read the vertical intercept at horizontal coordinate zero, then convert the logged parameters back to the original scale.

Both logarithmic axes require positive logged values. Plotting $\log y$ against $x$ tests an exponential form; plotting it against $\log x$ tests a power form. Interchanging these axes changes the model and the meaning of the gradient.

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—P3.4 - DifferentiationRead

P3.3 - Exponentials and logarithms

Learning objectives

Read every feature of a transformed exponential graph

Use natural logarithms to undo exponentials

Linearise a model to estimate its parameters

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