Investigating logarithmic functions

Exponential and logarithmic functions, grade 12 (4 units) Lesson 14 of 20Learning

Investigating logarithmic functions

The domain of

\displaystyle \ln f(x)

$\displaystyle \ln f(x)$ is defined only where $\displaystyle f(x)>0$ , so the domain is found by solving an inequality:

$\displaystyle \ln(x-2)$ : $\displaystyle x-2>0$ , that is $\displaystyle x>2$ .
$\displaystyle \ln(x^2-9)$ : $\displaystyle x^2-9>0$ , that is $\displaystyle x<-3$ or $\displaystyle x>3$ .
$\displaystyle \ln(4-x^2)$ : $\displaystyle 4-x^2>0$ , that is $\displaystyle -2<x<2$ .
In a quotient, the denominator is not 0: $\displaystyle \frac{1}{\ln x}$ needs $\displaystyle x>0$ and $\displaystyle \ln x\neq0$ , that is $\displaystyle x>0$ and $\displaystyle x\neq1$ .

A vertical asymptote

As $\displaystyle f(x)$ comes close to 0, $\displaystyle \ln f(x)$ falls without bound. So $\displaystyle \ln f(x)$ has a vertical asymptote at each edge of the domain where $\displaystyle {f(x)}={0}$ :

$\displaystyle \ln x$ : $\displaystyle {x}={0}$ ; $\displaystyle \ln(x-2)$ : $\displaystyle {x}={2}$ .
$\displaystyle \ln(x^2-9)$ has two vertical asymptotes: $\displaystyle {x}={-3}$ and $\displaystyle {x}={3}$ .

\displaystyle f(x)=\ln(x^{2}-9)

\displaystyle x=-3

\displaystyle x=3

A quotient has a vertical asymptote where its denominator is 0 and its numerator is not: in $\displaystyle \frac{2x}{\ln x-2}$ the denominator is 0 where $\displaystyle {\ln x}={2}$ , that is $\displaystyle {x}={e^2}$ , and the numerator there is $\displaystyle 2e^2\neq0$ . So $\displaystyle {x}={e^2}$ is a vertical asymptote.

Investigating

\displaystyle (\ln x)^2-2\ln x

The example below investigates the function $\displaystyle {f(x)}={(\ln x)^2-2\ln x}$ :

The domain: $\displaystyle x>0$ .
The x-axis: $\displaystyle {\ln x(\ln x-2)}={0}$ , so $\displaystyle {\ln x}={0}$ or $\displaystyle {\ln x}={2}$ : $\displaystyle {x}={1}$ or $\displaystyle {x}={e^2}$ . The graph does not meet the y-axis, since $\displaystyle {x}={0}$ is not in the domain.
The derivative: $\displaystyle {f'(x)}={\frac{2\ln x}{x}-\frac{2}{x}}={\frac{2\ln x-2}{x}}$ , and $\displaystyle {f'(x)}={0}$ where $\displaystyle {\ln x}={1}$ , that is $\displaystyle {x}={e}$ .
The denominator $\displaystyle x$ is positive in the domain, so the sign of $\displaystyle f'$ is the sign of $\displaystyle 2\ln x-2$ : decreasing where $\displaystyle 0<x<e$ , increasing where $\displaystyle x>e$ .
A minimum point: $\displaystyle {f(e)}={1-2}={-1}$ , that is $\displaystyle (e,-1)$ .
Positive where $\displaystyle 0<x<1$ and where $\displaystyle x>e^2$ , negative where $\displaystyle 1<x<e^2$ .
As $\displaystyle x$ comes close to 0, $\displaystyle \ln x$ falls without bound and $\displaystyle f(x)$ rises without bound: $\displaystyle {x}={0}$ is a vertical asymptote.

\displaystyle f(x)=(\ln x)^{2}-2\ln x

\displaystyle x=0

\displaystyle M=(2.718,-1)

An even function and an odd function

An even function satisfies $\displaystyle {f(-x)}={f(x)}$ for every $\displaystyle x$ in its domain, and its graph is symmetric about the y-axis. An odd function satisfies $\displaystyle {f(-x)}={-f(x)}$ , and its graph is symmetric about the origin.

The proof below shows that $\displaystyle {f(x)}={\ln(x^2-9)}$ is even: its domain, $\displaystyle x<-3$ or $\displaystyle x>3$ , is symmetric about 0, and for every $\displaystyle x$ in it

f(-x)=\ln\left((-x)^2-9\right)=\ln(x^2-9)=f(x)

A parameter from the slope of a tangent

The function $\displaystyle {f(x)}={2\ln(x^2-a)}$ is given, $\displaystyle a$ a positive parameter.
The slope of the tangent to the graph at the point where $\displaystyle {x}={5}$ is 1.25.

The example below finds $\displaystyle a$ and investigates the function:

$\displaystyle {f'(x)}={\frac{2\cdot2x}{x^2-a}}={\frac{4x}{x^2-a}}$ , so $\displaystyle {f'(5)}={\frac{20}{25-a}}$ .
$\displaystyle {\frac{20}{25-a}}={1.25}$ : $\displaystyle {25-a}={16}$ , so $\displaystyle {a}={9}$ and $\displaystyle {f(x)}={2\ln(x^2-9)}$ .
The domain: $\displaystyle x<-3$ or $\displaystyle x>3$ ; the vertical asymptotes: $\displaystyle {x}={-3}$ and $\displaystyle {x}={3}$ .
$\displaystyle {f'(x)}={\frac{4x}{x^2-9}}$ : where $\displaystyle x>3$ , $\displaystyle f'(x)>0$ – increasing; where $\displaystyle x<-3$ , $\displaystyle f'(x)<0$ – decreasing. There are no extremum points.
The x-axis: $\displaystyle {x^2-9}={1}$ , so $\displaystyle {x}={\sqrt{10}}$ or $\displaystyle {x}={-\sqrt{10}}$ . The graph does not meet the y-axis.
The function is even, like $\displaystyle \ln(x^2-9)$ .
The function $\displaystyle {g(x)}={f(x-4)}$ is $\displaystyle f$ shifted 4 to the right: its asymptotes are $\displaystyle {x}={1}$ and $\displaystyle {x}={7}$ , and its graph is symmetric about the line $\displaystyle {x}={4}$ – not about the y-axis. $\displaystyle g$ is neither even nor odd.

\displaystyle f(x)=2\ln(x^{2}-9)

\displaystyle x=-3

\displaystyle x=3

Exercise 1

Find the domain of each function: write an inequality and solve it.

$\displaystyle {f(x)}={\ln(x+4)}$

$\displaystyle {f(x)}={\ln(x^2-4x)}$

$\displaystyle {f(x)}={\ln(6-x-x^2)}$

Exercise 2

The function $\displaystyle {f(x)}={\frac{\ln x}{x}}$ is given.

Find $\displaystyle f'(x)$ .

$\displaystyle f(x)=\frac{\ln x}{x}$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Find the interval of increase: solve the inequality $\displaystyle f'(x)>0$ .

Mark on the graph the extremum point and the point where it meets the x-axis.

\displaystyle f(x)=\frac{\ln x}{x}

Write the coordinates of the extremum point and its kind, and the equation of the vertical asymptote.

Exercise 3

The function $\displaystyle {f(x)}={\ln(x^2+a)}$ is given, $\displaystyle a$ a positive parameter.
The slope of the tangent to the graph at the point where $\displaystyle {x}={1}$ is 1.

Find $\displaystyle a$ .

Put in the $\displaystyle a$ you found.
Mark the extremum point on the graph.

\displaystyle f(x)=\ln(x^{2}+1)

Prove that $\displaystyle f$ is even, and find its domain.

Exercise 4

The function $\displaystyle {f(x)}={\frac{2x}{\ln x-a}}$ is given, $\displaystyle a$ a positive parameter.
The line $\displaystyle {y}={2x}$ meets the graph of $\displaystyle f$ at the point where $\displaystyle {x}={e^3}$ .

Find $\displaystyle a$ .

Put in $\displaystyle {a}={2}$ .
Find the domain of $\displaystyle f$ and the equation of its vertical asymptote.

Find $\displaystyle f'(x)$ .

$\displaystyle f(x)=\frac{2x}{\ln x-2}$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Solve the equation $\displaystyle {f'(x)}={0}$ .

Mark the extremum point on the graph.

\displaystyle f(x)=\frac{2x}{\ln x-2}

\displaystyle x=e^{2}

Formulas: exponential and logarithmic functions, grade 12 (4 units)

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Powers

A product of powers with the same base
$\displaystyle a^{x}\cdot a^{y}$ $\displaystyle =$ $\displaystyle a^{x+y}$
A quotient of powers with the same base
$\displaystyle \frac{a^{x}}{a^{y}}$ $\displaystyle =$ $\displaystyle a^{x-y}$
A power of a power
$\displaystyle (a^{x})^{y}$ $\displaystyle =$ $\displaystyle a^{x\cdot y}$
A power of a product
$\displaystyle (a\cdot b)^{x}$ $\displaystyle =$ $\displaystyle a^{x}\cdot b^{x}$
A power of a quotient
$\displaystyle \left(\frac{a}{b}\right)^{x}$ $\displaystyle =$ $\displaystyle \frac{a^{x}}{b^{x}}$
A negative exponent
$\displaystyle a^{-x}$ $\displaystyle =$ $\displaystyle \frac{1}{a^{x}}$
The exponent 0
$\displaystyle a^{0}$ $\displaystyle =$ $\displaystyle 1$ $\displaystyle a\neq0$
A rational exponent
$\displaystyle a^{\frac{m}{n}}$ $\displaystyle =$ $\displaystyle \sqrt[n]{a^{m}}$ $\displaystyle a>0$

Logarithms (within the domain)

The definition of the logarithm
$\displaystyle \log_{a}x=b\iff a^{b}=x$ $\displaystyle a>0,\ a\neq1,\ x>0$
$\displaystyle \log_{a}(a^{b})$ $\displaystyle =$ $\displaystyle b$
$\displaystyle a^{\log_{a}x}$ $\displaystyle =$ $\displaystyle x$
The logarithm of a product
$\displaystyle \log_{a}x+\log_{a}y$ $\displaystyle =$ $\displaystyle \log_{a}(x\cdot y)$
The logarithm of a quotient
$\displaystyle \log_{a}x-\log_{a}y$ $\displaystyle =$ $\displaystyle \log_{a}\left(\frac{x}{y}\right)$
The logarithm of a power
$\displaystyle \log_{a}(x^{b})$ $\displaystyle =$ $\displaystyle b\cdot\log_{a}x$
The natural logarithm – to the base $\displaystyle e$
$\displaystyle \ln x=\log_{e}x$

Growth and decay

The amount after $\displaystyle t$ units of time
$\displaystyle f(t)=f(0)\cdot q^{t}$
The growth / decay factor – $\displaystyle p$ the percent per unit of time
$\displaystyle q=1\pm\frac{p}{100}$
The half-life $\displaystyle T$
$\displaystyle q^{T}$ $\displaystyle =$ $\displaystyle \frac{1}{2}$

Derivatives

$\displaystyle (e^{x})^{\prime}$ $\displaystyle =$ $\displaystyle e^{x}$
$\displaystyle (\ln x)^{\prime}$ $\displaystyle =$ $\displaystyle \frac{1}{x}$
The derivative of $\displaystyle e^{f(x)}$
$\displaystyle \left[e^{f(x)}\right]^{\prime}$ $\displaystyle =$ $\displaystyle e^{f(x)}\cdot f^{\prime}(x)$
The derivative of $\displaystyle \ln(f(x))$
$\displaystyle \left[\ln(f(x))\right]^{\prime}$ $\displaystyle =$ $\displaystyle \frac{f^{\prime}(x)}{f(x)}$
The derivative of a product
$\displaystyle \left[f(x)\cdot g(x)\right]^{\prime}$ $\displaystyle =$ $\displaystyle f^{\prime}(x)\cdot g(x)+f(x)\cdot g^{\prime}(x)$
The derivative of a quotient
$\displaystyle \left[\frac{f(x)}{g(x)}\right]^{\prime}$ $\displaystyle =$ $\displaystyle \frac{f^{\prime}(x)\cdot g(x)-f(x)\cdot g^{\prime}(x)}{\left[g(x)\right]^{2}}$
The derivative of a composite function
$\displaystyle \left[f(u(x))\right]^{\prime}$ $\displaystyle =$ $\displaystyle f^{\prime}(u)\cdot u^{\prime}(x)$
The tangent’s equation at the point where $\displaystyle {x}={a}$
$\displaystyle y-f(a)=f^{\prime}(a)(x-a)$

Integrals and the area

$\displaystyle \int e^{x}\,dx=e^{x}+C$
$\displaystyle \int e^{mx+b}\,dx=\frac{e^{mx+b}}{m}+C$ $\displaystyle m\neq0$
In any interval where $\displaystyle x\neq0$
$\displaystyle \int\frac{1}{x}\,dx=\ln|x|+C$
$\displaystyle \int\frac{1}{mx+b}\,dx=\frac{\ln|mx+b|}{m}+C$ $\displaystyle m\neq0$
$\displaystyle F$ an antiderivative of $\displaystyle f$
$\displaystyle \int_{a}^{b}f(x)\,dx=F(b)-F(a)$
The area between two graphs
$\displaystyle S=\int_{a}^{b}\left(f(x)-g(x)\right)dx$ $\displaystyle f(x)\geq g(x)$