Logarithmic and exponential functions

Unit 2 Logarithmic and exponential functions

Key unit competence: Extend the concepts of functions to investigate logarithmic and exponential functions and their applications.

Key ideas

• $\ln x$ is defined on $]0,+\infty [$ with range $\mathbb{R}$; $\ln 1=0$, $\ln e=1$; $\ln (ab)=\ln a+\ln b$, $\ln \frac{a}{b}=\ln a-\ln b$, $\ln a^n=n\ln a$.
• Limits: ${\lim }_{x\to 0^{+}}\ln x=-\infty$ (vertical asymptote $x=0$), ${\lim }_{x\to +\infty }\ln x=+\infty$, ${\lim }_{x\to +\infty }\frac{\ln x}{x}=0$, ${\lim }_{x\to 0^{+}}x\ln x=0$.
• Derivatives: $(\ln u{)}^{\prime }=\frac{u^{\prime }}{u}$; $({\log }_{a}u{)}^{\prime }=\frac{u^{\prime }}{u\ln a}$ where ${\log }_{a}x=\frac{\ln x}{\ln a}$.
• Logarithmic differentiation: for products, quotients and powers like $x^x$, take $\ln$ of both sides, then differentiate implicitly.
• $e^x$ is the inverse of $\ln x$: domain $\mathbb{R}$, range $]0,+\infty [$; ${\lim }_{x\to -\infty }{e}^x=0$ (horizontal asymptote $y=0$), ${\lim }_{x\to +\infty }\frac{e^x}{x}=+\infty$. $(e^u{)}^{\prime }=u^{\prime }{e}^u$.
• $a^x=e^{x\ln a}$: increasing if $a>1$, decreasing if $0<a<1$; $(a^u{)}^{\prime }=u^{\prime }{a}^u\ln a$.
• Applications: compound interest $A=P{\left(1+\frac{r}{n}\right)}^{nt}$, continuous $A=P{e}^{rt}$; growth $P=P_0{e}^{rt}$; depreciation $V=V_0(1-r{)}^t$; Richter magnitude $M=\log \frac{x}{x_0}$; carbon-14 half-life about 5700 years.

Formulas to know

Common misconceptions

• "$\ln (a+b)=\ln a+\ln b$." No: $\ln (ab)=\ln a+\ln b$.
• "$(3^x{)}^{\prime }=x{3}^{x-1}$." No: that rule is for $x^n$; $(3^x{)}^{\prime }=3^x\ln 3$.
• Forgetting to check the domain: solutions of log equations must make every logarithm defined.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Study the function $f(x)=\frac{\ln x}{x}$: domain, limits, asymptotes, derivative and maximum.

   Show answer

   Domain $]0,+\infty [$. ${\lim }_{x\to 0^{+}}f=-\infty$ (VA $x=0$); ${\lim }_{x\to +\infty }f=0$ (HA $y=0$). $f^{\prime }(x)=\frac{1-\ln x}{x^2}$: positive for $x<e$, negative for $x>e$. Maximum $f(e)=\frac{1}{e}$.

   Common mistake: Forgetting the horizontal asymptote $y=0$.

2. Find ${\lim }_{x\to +\infty }(x-\ln x)$.

   Show answer

   $x-\ln x=x\left(1-\frac{\ln x}{x}\right)$ and $\frac{\ln x}{x}\to 0$, so the limit is $+\infty$.

   Common mistake: Calling it the indeterminate form $\infty -\infty$ and stopping.

3. Differentiate $y={\log }_3(x^2+1)$.

   Show answer

   $y^{\prime }=\frac{2x}{(x^2+1)\ln 3}$.

   Common mistake: Omitting the factor $\frac{1}{\ln 3}$.

4. A car worth 12 000 000 FRW depreciates by 15% per year. Find its value after 4 years.

   Show answer

   $12000000\times 0.85^4\approx 6264075$ FRW.

   Common mistake: Subtracting $4\times 15\%=60\%$.

5. Solve $\ln (x+1)+\ln (x-1)=\ln 3$.

   Show answer

   Domain $x>1$. $\ln (x^2-1)=\ln 3$, so $x^2=4$; $x=2$ ($x=-2$ is rejected).

   Common mistake: Keeping $x=-2$.

6. A population of 10 million grows by 2.5% per year. After how many years does it double?

   Show answer

   $1.025^n=2$, so $n=\frac{\ln 2}{\ln 1.025}\approx 28.07$: about 28 years.

   Common mistake: Using simple growth: $\frac{100}{2.5}=40$ years.

7. Use logarithmic differentiation to find $y^{\prime }$ for $y=\frac{(x+1)(x-2{)}^3}{x-3}$.

   Show answer

   $\ln y=\ln (x+1)+3\ln (x-2)-\ln (x-3)$, so $y^{\prime }=\frac{(x+1)(x-2{)}^3}{x-3}\left(\frac{1}{x+1}+\frac{3}{x-2}-\frac{1}{x-3}\right)$.

   Common mistake: Forgetting to multiply back by $y$.

8. A mortgage of 10 000 000 FRW is repaid monthly over 5 years at 12% per year. Find the monthly payment using $P=\frac{Mi}{1-(1+i{)}^{-n}}$.

   Show answer

   $i=0.01$, $n=60$: $P=\frac{10000000\times 0.01}{1-1.01^{-60}}\approx 222444$ FRW.

   Common mistake: Using $i=0.12$ instead of the monthly rate.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Differentiate $y=\ln (3x^2+1)$.	a. $\frac{1}{e}$ at $x=1$
   2. Find ${\lim }_{x\to +\infty }{\left(1+\frac{2}{x}\right)}^x$.	b. $\frac{6x}{3x^2+1}$
   3. Find the inverse of $f(x)=e^{2x}+1$.	c. $f^{-1}(x)=\frac{1}{2}\ln (x-1)$, $x>1$
   4. Find the maximum value of $f(x)=x{e}^{-x}$.	d. $e^2$

   Show answer

   1 → b ($\frac{6x}{3x^2+1}$) · 2 → d ($e^2$) · 3 → c ($f^{-1}(x)=\frac{1}{2}\ln (x-1)$, $x>1$) · 4 → a ($\frac{1}{e}$ at $x=1$)

   • Step 1. $(\ln u{)}^{\prime }=\frac{u^{\prime }}{u}$ with $u=3x^2+1$.
     Step 2. $u^{\prime }=6x$, so $y^{\prime }=\frac{6x}{3x^2+1}$.
   • Step 1. $\ln {\left(1+\frac{2}{x}\right)}^x=x\ln \left(1+\frac{2}{x}\right)\approx x\cdot \frac{2}{x}=2$.
     Step 2. So the limit is $e^2$.
   • Step 1. $y=e^{2x}+1\Rightarrow e^{2x}=y-1\Rightarrow x=\frac{1}{2}\ln (y-1)$.
     Step 2. Swap letters: $f^{-1}(x)=\frac{1}{2}\ln (x-1)$ for $x>1$.
   • Step 1. $f^{\prime }(x)=e^{-x}(1-x)=0\Rightarrow x=1$.
     Step 2. $f^{\prime }$ changes from $+$ to $-$, so maximum $f(1)=\frac{1}{e}$.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   11 400 years  ·  $\frac{2\ln 5}{\ln 3}\approx 2.93$ h  ·  $-\frac{1}{e}$  ·  0
   1. Bacteria follow $N=N_0{e}^{kt}$ and triple in 2 hours. They become 5 times the initial number after .
   2. A sample has 25% of its original carbon-14 (half-life 5700 years). Its age is about .
   3. The minimum value of $f(x)=x\ln x$ is .
   Show answer

   1: $\frac{2\ln 5}{\ln 3}\approx 2.93$ h · 2: 11 400 years · 3: $-\frac{1}{e}$

   • Step 1. $e^{2k}=3\Rightarrow k=\frac{\ln 3}{2}$.
     Step 2. $e^{kt}=5\Rightarrow t=\frac{\ln 5}{k}=\frac{2\ln 5}{\ln 3}\approx 2.93$ h.
   • Step 1. $25\%={\left(\frac{1}{2}\right)}^2$: two half-lives have passed.
     Step 2. $2\times 5700=11400$ years.
   • Step 1. $f^{\prime }(x)=\ln x+1=0\Rightarrow x=\frac{1}{e}$; $f^{\prime }$ changes from $-$ to $+$.
     Step 2. $f\left(\frac{1}{e}\right)=\frac{1}{e}\ln \frac{1}{e}=-\frac{1}{e}$.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Find ${\lim }_{x\to 0}\frac{\ln (1+2x)}{x}$.	a. $e^{2x}(2\sin x+\cos x)$
   2. Solve $\ln (x+1)+\ln (x-1)=\ln 3$.	b. $x=2$
   3. Find the domain of $f(x)=\ln (x^2-4)$.	c. $]-\infty ,-2[\cup ]2,+\infty [$
   4. Differentiate $y=e^{2x}\sin x$.	d. $2$

   Show answer

   1 → d ($2$) · 2 → b ($x=2$) · 3 → c ($]-\infty ,-2[\cup ]2,+\infty [$) · 4 → a ($e^{2x}(2\sin x+\cos x)$)

   • Step 1. Write $\frac{\ln (1+2x)}{2x}\times 2$.
     Step 2. ${\lim }_{u\to 0}\frac{\ln (1+u)}{u}=1$, so the limit is 2.
   • Step 1. Domain: $x>1$. Combine: $\ln (x^2-1)=\ln 3$.
     Step 2. $x^2-1=3\Rightarrow x=\pm 2$; only $x=2$ is in the domain.
   • Step 1. Need $x^2-4>0$, i.e. $x^2>4$.
     Step 2. So $x<-2$ or $x>2$.
   • Step 1. Product rule: $(e^{2x}{)}^{\prime }\sin x+e^{2x}(\sin x{)}^{\prime }$.
     Step 2. $2e^{2x}\sin x+e^{2x}\cos x=e^{2x}(2\sin x+\cos x)$.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   4  ·  1 552 924 FRW  ·  $x^x(1+\ln x)$  ·  560 000 FRW
   1. Two earthquakes measure 8.9 and 8.3 on the Richter scale. The first is stronger by a factor of about .
   2. For $x>0$, the derivative of $y=x^x$ is .
   3. 500 000 FRW is invested at 12% compounded annually. After 10 years the amount is about .
   Show answer

   1: 4 · 2: $x^x(1+\ln x)$ · 3: 1 552 924 FRW

   • Step 1. Each unit on the Richter scale is a factor of 10: ratio $=10^{8.9-8.3}$.
     Step 2. $10^{0.6}\approx 3.98\approx 4$.
   • Step 1. $\ln y=x\ln x$.
     Step 2. Differentiate: $\frac{y^{\prime }}{y}=\ln x+1$, so $y^{\prime }=x^x(1+\ln x)$.
   • Step 1. $A=P(1+r{)}^n=500000\times 1.12^{10}$.
     Step 2. $1.12^{10}\approx 3.1058$, so $A\approx 1552924$ FRW.

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