\documentclass[12pt,oneside]{article}
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\usepackage{amssymb,amsmath,amsthm}
\usepackage[table,xcdraw]{xcolor}
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% looks normal. Note that 1.1=1/.9090909...
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\renewcommand{\arraystretch}{.91}
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% \newenvironment{exercise}[1]{\vspace{.5cm}\noindent\textbf{#1 \hspace{.05em}}}{}
% Allow for underlining.
\usepackage[normalem]{ulem}
% define shortcut commands for commonly used symbols
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\begin{document}
Unit 2 Assessment - Albert Xia
\vspace{6mm}
1. Fill in the table.
\begin{align*}
\begin{tabular}{|c|c|}
\hline
Exponential form & Logarithmic Form \\ \hline
$2401=7^4$ & $\log_7(2401)=4$ \\ \hline
$a^c =b$ & $a=\log_b c$ \\ \hline
\end{tabular}
\end{align*}
\vspace{6mm}
2. Solve for $x$. Give answer to three decimal places where applicable.
$$\text{a.}\quad x=\log37$$
\begin{align*}
10^x&=37\\
x&=1.568
\end{align*}
$$\text{b.}\quad x=\log10000$$
\begin{align*}
10^x&=10000\\
x&=4
\end{align*}
3. Convert each of the following to exponential form and solve.
$$\text{a.}\quad n=\log_416384$$
\begin{align*}
4^n&=16384\\
4^n&=4^7\\
n&=7
\end{align*}
$$\text{b.}\quad 6=\log_b46656$$
\begin{align*}
b^6&=46656\\
b&=\sqrt{6}46656\\
b&=6
\end{align*}
\newpage
4. Use the special properties of logarithms to solve the following equations.
$$\text{a.}\quad M=\log_{11}11^{21}$$
\qquad\text{Using} the property $log_{a}a^{x}=x$...
\begin{align*}
M&=21\log_{11}(11)\\
M&=21
\end{align*}
$$\text{b.}\quad 4^{\log_{4}57}=x$$
\qquad\text{Using} the property $a^{log_{a}x}=x$...
\begin{align*}
x&=4^{log_{4}57}\\
x&=57
\end{align*}
$$\text{c.}\quad x=\log_{23}78$$
\qquad Using the proprety $\log_{a}b=\frac{\log a}{\log b}$...
\begin{align*}
x=\frac{\log78}{\log23}\\
x\approx1.389
\end{align*}
\vspace{6mm}
5. Evaluate the following using a method of your choice.
$$\text{a.}\quad\log_9729$$
\qquad Let $x$ be the unknown variable.
\begin{align*}
7^x&=729\\
x&=3.387\\
\end{align*}
$$\text{b.}\quad\log_{6}7776-\log_{6}36$$
\qquad$\log_{6}7776\text{ can be rewritten as }6^n=7776$
\begin{align*}
6^n&=7776\\
6^n&=6^5\\
n&=5
\end{align*}
\qquad$\log_{6}36\text{ can be rewritten as }6^x=36$
\begin{align*}
6^x&=36\\
6^x&=6^2\\
x&=2
\end{align*}
\qquad Now we can substitute $\log_{6}7776-\log_{6}36$ with $n-x$
\begin{align*}
\log_{6}7776-\log_{6}36&=n-x\\
n-x&=5-2\\
n-x&=3
\end{align*}
\vspace{6mm}
6. Solve for the unknown variable using a method of your choice.
$$\text{a.}\quad5^{11x+23}=125^{7x}$$
\qquad I will solve for the unknown variable by making the bases equal.
\begin{align*}
5^{11x+23}&=((5)^3)^{7x}\\
5^{11x+23}&=(5)^21x\\
11x+23&=21x\\
23&=21x-11x\\
23&=10x\\
\frac{23}{10}&=x
\end{align*}
$$\text{b.}\quad2^n=123$$
\qquad I will solve for the unknown variable using the property $q\log_b p=\log_b p^q$.
\begin{align*}
2^n&=123\\
\log2^n&=\log123\\
n\log2&=log123\\
n&=\frac{\log123}{\log2}\\
n&\approx6.943
\end{align*}
\newpage
7. Solve for the unknown variable using a method of your choice.
\begin{align*}
3^{x+5}+3^x&=177876\\
(3^x)(3^5)+3^x&=177876\\
3^x(3^5+1)&=177876\\
3^x(244)&=177876\\
3^x&=\frac{177876}{244}\\
3^x&=729\\
3^x&=3^6\\
x&=6
\end{align*}
\vspace{6mm}
8. Describe the transformations on the given logarithmic function.
$$g(x)=-\frac{1}{6}\log_{10}[-\frac{1}{6}(x-7)]-85$$
\vspace{2mm}
The transformations on the given logarithmic function is as follows:
\begin{enumerate}
\item The parent function of this function is $g(x)=a\log_{10}x$.
\item The transformed function is reflected in the x-axis.
\item The transformed function is being vertically compressed by a factor of $\frac{1}{6}$.
\item The transformed function is being reflected in the y-axis.
\item The transformed function is being horizontally stretched by a factor of 6.
\item The transformed function is being translated right 7 units.
\item The transformed function is being translated down 85 units.
\end{enumerate}
\vspace{6mm}
9. Write the equation of a logarithmic function that has been vertically stretched by 5, reflected in the x-axis, horizontally compressed by $\frac{1}{3}$, moved 11 units left and 8 units down. Justify your answer.
\vspace{3mm}
I define the transformed function as $f(x)$.
\begin{align*}
a=-5\\
k=3\\
d=-11\\
c=-8
\end{align*}
The transformed function is $f(x)=-5\log_{10}[3(x+11)]-8$.
\vspace{12mm}
10. The point $(-23,\,-13)$ is on the transformed function: $y=-2\log_{10}[-5(x+3)]-9$. Find the original point on the parent function $f(x)=\log_{10}x$. Justify your answer.
\vspace{3mm}
\begin{align*}
a&=-2\\
k&=-5\\
d&=-3\\
c&=-9
\end{align*}
I will use mapping rule to determine the x and y values translated onto the parent function.
$$(x,y)\rightarrow(-23,\,-13)$$
$$(x,y)\rightarrow(\frac{x}{k}+d,\,ay+c)$$
\vspace{3mm}
So we have for the $x$ value...
\begin{align*}
\frac{x}{-5}-3&=-23\\
\frac{x}{-5}&=-20\\
x&=-20(-5)\\
x&=100
\end{align*}
And for the $y$ value...
\begin{align*}
-2(y)+(-9)&=-13\\
-2(y)&=(-13)+9\\
-2(y)&=-4\\
y&=\frac{-4}{-2}\\
y&=2
\end{align*}
Therefore we determine the original point on the parent function $f(x)=\log_{10}x$ to be $(100,2)$.
\vspace{6mm}
11. Ethan bought a new car worth 60000 dollars. After 5 years, the car was worth 35429.40 dollars. Calculate the depreciation rate of Ethan's car.
\newpage
The equation I will use is $W=C(1-\frac{D}{100})^T$ where $C$ is initial cost, $W$ is worth of car, $D$ is rate of depreciation, and $T$ is time in years.
\begin{align*}
W&=C(1-\frac{D}{100})^T\\
35429.40&=60000(1-\frac{D}{100})^5\\
\frac{35429.40}{60000}&=(1-\frac{D}{100})^5\\
(1-\frac{D}{100})&=(\frac{35429.40}{60000})^{\frac{1}{5}}\\
(\frac{35429.40}{60000})^{\frac{1}{5}}&=0.9\\
0.9&=(1-\frac{D}{100})\\
\frac{D}{100}&=1-0.9\\
\frac{D}{100}&=0.1\\
D&=0.1*100\\
D&=10\text{ (percent)}
\end{align*}
The depreciation rate of Ethan's car is 10 percent.
\vspace{12mm}
12. The hydrogen ion concentration of a lemon is about 0.01. What is its pH?
\vspace{6mm}
I will convert the value of the lemon's hydrogen ion concentration 0.01 into a base ten number for the logarithmic equation, which is $10^{-2}$.
\vspace{6mm}
I will determine the pH by using the equation $pH=-\log[H^+]$, where the value of $[H^+]$ is $10^{-2}\,mol/l$.
\begin{align*}
pH&=-\log(10^{-2})\\
pH&=-2\log_{10}(10)\\
pH&=-(-2)\\
pH&=2
\end{align*}
The pH of a lemon is 2.
\vspace{12mm}
13. How does an earthquake of magnitude 8.4 compare to an earthquake of magnitude 3.7? Round answer to the nearest whole number.
To calculate the difference in size, I will divide $10^{8.4}$ from $10^{3.7}$ since the Richter scale is logarithmic.
\newpage
\begin{align*}
\text{Size_difference}&=\frac{10^{8.4}}{10^{3.7}}\\
\text{Size_difference}&=10^{8.4-3.7}\\
\text{Size_difference}&=10^{4.7}\\
\text{Size_difference}&\approx50119
\end{align*}
An earthquake of 8.4 magnitude on the Richter scale is approximately 50119 times larger than one of 3.7 magnitude on the same scale.
\end{document}
答案1
将上一个 align*-environment 更改为
\begin{align*}
\text{Size\_difference}&=\frac{10^{8.4}}{10^{3.7}}\\
\text{Size\_difference}&=10^{8.4-3.7}\\
\text{Size\_difference}&=10^{4.7}\\
\text{Size\_difference}&\approx50119
\end{align*}
应该可以成功编译整个手稿。
答案2
除了修复语法错误（忘记转义文字_（下划线）字符）之外，您可能还需要熟悉避免执行过多视觉格式化的方法。
我附上了一些建议，让你的文档“表现”得像 LaTeX 文档。枚举列表中的最后一项将如下所示：
\documentclass[12pt,oneside]{article}
%\setlength{\parindent}{0pt} % not needed
% This package simply sets the margins to be 1 inch.
\usepackage[margin=1in]{geometry}
\usepackage{array}
\usepackage{graphicx}
% This removes page numbers.
\pagestyle{empty} %%% \pagenumbering{gobble}
% These packages include nice commands from AMS-LaTeX
\usepackage{amssymb,amsmath,amsthm}
\usepackage[table,xcdraw]{xcolor}
%% Make the space between lines slightly more
%% generous than normal single spacing, but compensate
%% so that the spacing between rows of matrices still
%% looks normal. Note that 1.1=1/.9090909...
%%\renewcommand{\baselinestretch}{1.1}
%%\renewcommand{\arraystretch}{.91}
% The preceding reasoning is incorrect! Better to use the setspace package:
\usepackage[nodisplayskipstretch]{setspace}
\setstretch{1.1}
% Define an environment for exercises.
% \newenvironment{exercise}[1]{\vspace{.5cm}\noindent\textbf{#1 \hspace{.05em}}}{}
% Allow for underlining.
\usepackage[normalem]{ulem}
% define shortcut commands for commonly used symbols
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\calP}{\mathcal{P}}
\DeclareMathOperator{\vsspan}{span}
%% New code:
\usepackage[T1]{fontenc}
\usepackage{enumitem}
\setlist[enumerate,1]{label=\bfseries\arabic*.,left=0pt}
\setlist[enumerate,2]{label=\alph*.,left=0pt}
\usepackage[print-unity-mantissa=false,per-mode=symbol]{siunitx}
\allowdisplaybreaks
\begin{document}
\subsubsection*{Unit 2 Assessment --- Albert Xia}
%%% \vspace{6mm}
\begin{enumerate}
% #1
\item Fill in the table.
%%%%%\begin{align*}
\begin{center}
\setlength{\extrarowheight}{2pt}
\begin{tabular}{|c|c|}
\hline
Exponential form & Logarithmic Form \\ \hline
$2401=7^4$ & $\log_7(2401)=4$ \\ \hline
$a^c =b$ & $a=\log_b c$ \\ \hline
\end{tabular}
%%%%%\end{align*}
\end{center}
%%% \vspace{6mm}
% #2
\item Solve for $x$. Give answer to three decimal places where applicable.
\begin{enumerate}
\item $x=\log37$
\begin{align*}
10^x&=37\\
x&=1.568
\end{align*}
\item $x=\log10000$
\begin{align*}
10^x&=10000\\
x&=4
\end{align*}
\end{enumerate}
% #3
\item Convert each of the following to exponential form and solve.
\begin{enumerate}
\item $n=\log_416384$
\begin{align*}
4^n&=16384\\
4^n&=4^7\\
n&=7
\end{align*}
\item $6=\log_b46656$
\begin{align*}
b^6&=46656\\
b&=\sqrt[6]{46656}\\
b&=6
\end{align*}
\end{enumerate}
% #4
\item Use the special properties of logarithms to solve the following equations.
\begin{enumerate}
\item $M=\log_{11}11^{21}$
Using the property $\log_{a}a^{x}=x$:
\begin{align*}
M&=21\log_{11}(11)\\
&=21
\end{align*}
\item $4^{\log_{4}57}=x$
Using the property $a^{\log_{a}x}=x$:
\begin{align*}
x&=4^{\log_{4}57}\\
x&=57
\end{align*}
\item $x=\log_{23}78$
Using the proprety $\log_{a}b=\frac{\log a}{\log b}$:
\begin{align*}
x&=\frac{\log78}{\log23}\\
&\approx1.389
\end{align*}
\end{enumerate}
% #5
\item Evaluate the following using a method of your choice.
\begin{enumerate}
\item $\log_9729$.
Let $x$ be the unknown variable.
\begin{align*}
7^x&=729\\
x&=3.387
\end{align*}
\item $\log_{6}7776-\log_{6}36$.
$\log_{6}7776$ can be rewritten as $6^n=7776$
\begin{align*}
6^n&=7776\\
6^n&=6^5\\
n&=5
\end{align*}
$\log_{6}36$ can be rewritten as $6^x=36$
\begin{align*}
6^x&=36\\
6^x&=6^2\\
x&=2
\end{align*}
Now we can substitute $\log_{6}7776-\log_{6}36$ with $n-x$
\begin{align*}
\log_{6}7776-\log_{6}36&=n-x\\
n-x&=5-2\\
n-x&=3
\end{align*}
\end{enumerate}
\newpage
% #6
\item Solve for the unknown variable using a method of your choice.
\begin{enumerate}
\item $5^{11x+23}=125^{7x}$
I will solve for the unknown variable $x$ by making the bases equal.
\begin{align*}
5^{11x+23}&=((5)^3)^{7x}\\
5^{11x+23}&=(5)^{21x}\\
11x+23&=21x\\
23&=21x-11x\\
23&=10x\\
\frac{23}{10}&=x
\end{align*}
\item $2^n=123$
I will solve for the unknown variable using the property $q\log_b p=\log_b p^q$.
\begin{align*}
2^n&=123\\
\log2^n&=\log123\\
n\log2&=\log123\\
n&=\frac{\log123}{\log2}\\
n&\approx6.943
\end{align*}
\end{enumerate}
% #7
\item Solve for the unknown variable using a method of your choice.
\begin{align*}
3^{x+5}+3^x&=177876\\
(3^x)(3^5)+3^x&=177876\\
3^x(3^5+1)&=177876\\
3^x(244)&=177876\\
3^x&=\frac{177876}{244}\\
3^x&=729\\
3^x&=3^6\\
x&=6
\end{align*}
%%% \vspace{6mm}
% #8
\item Describe the transformations on the given logarithmic function.
\[
g(x)=-\frac{1}{6}\log_{10}\Bigl[-\frac{1}{6}(x-7)\Bigr]-85
\]
%\vspace{2mm}
The transformations on the given logarithmic function are as follows:
\begin{enumerate}
\item The parent function of this function is $g(x)=a\log_{10}x-c$.
\item The transformed function is being reflected in the $x$-axis.
\item The transformed function is being vertically compressed by a factor of $\frac{1}{6}$.
\item The transformed function is being reflected in the $y$-axis.
\item The transformed function is being horizontally stretched by a factor of~6.
\item The transformed function is being translated right 7 units.
\item The transformed function is being translated down 85 units.
\end{enumerate}
%%% \vspace{6mm}
% #9
\item Write the equation of a logarithmic function that has been vertically stretched by 5, reflected in the x-axis, horizontally compressed by $\frac{1}{3}$, moved 11 units left and 8 units down. Justify your answer.
%\vspace{3mm}
I define the transformed function as $f(x)$.
\begin{align*}
a&=-5\\
k&=3\\
d&=-11\\
c&=-8
\end{align*}
The transformed function is $f(x)=-5\log_{10}[3(x+11)]-8$.
%\vspace{12mm}
% #10
\item The point $(-23,-13)$ is on the transformed function: $y=-2\log_{10}[-5(x+3)]-9$. Find the original point on the parent function $f(x)=\log_{10}x$. Justify your answer.
%\vspace{3mm}
\begin{align*}
a&=-2\\
k&=-5\\
d&=-3\\
c&=-9
\end{align*}
I will use mapping rule to determine the $x$ and $y$ values translated onto the parent function.
\begin{align*}
(x,y)&\rightarrow(-23,\,-13)\\
(x,y)&\rightarrow\Bigl(\frac{x}{k}+d,\,ay+c\Bigr)
\end{align*}
%\vspace{3mm}
\enlargethispage{1.5\baselineskip} % only instance of visual formatting
So we have for the $x$ value
\begin{align*}
\frac{x}{-5}-3&=-23\\
\frac{x}{-5}&=-20\\
x&=-20(-5)\\
x&=100
\end{align*}
And for the $y$ value...
\begin{align*}
-2(y)+(-9)&=-13\\
-2(y)&=(-13)+9\\
-2(y)&=-4\\
y&=\frac{-4}{-2}\\
y&=2
\end{align*}
Therefore we determine the original point on the parent function $f(x)=\log_{10}x$ to be $(100,2)$.
%%% \vspace{6mm}
% #11
\item Ethan bought a new car worth \$60000. After 5 years, the car was worth \$35429.40. Calculate the depreciation rate of Ethan's car.
%\newpage
The equation I will use is $W=C\bigl(1-\frac{D}{100}\bigr)^T$, where $C$ is initial cost (in dollars), $W$ is the current worth of the car (in dollars), $D$ is the annualized rate of depreciation, and $T$ is time (in years).
\begin{align*}
W&=C\Bigl(1-\frac{D}{100}\Bigr)^T\\
35429.40&=60000\Bigl(1-\frac{D}{100}\Bigr)^5\\
\frac{35429.40}{60000}&=\Bigl(1-\frac{D}{100}\Bigr)^5\\
\Bigl(1-\frac{D}{100}\Bigr)&=\Bigl(\frac{35429.40}{60000}\Bigr)^{\frac{1}{5}}\\
\Bigl(\frac{35429.40}{60000}\Bigr)^{\frac{1}{5}}&=0.9\\
0.9&=\Bigl(1-\frac{D}{100}\Bigr)\\
\frac{D}{100}&=1-0.9\\
\frac{D}{100}&=0.1\\
D&=0.1\cdot100\\
D&=10\%\,.
\end{align*}
The depreciation rate of Ethan's car is 10 percent.
%\vspace{12mm}
% #12
\item The hydrogen ion concentration of a lemon is about~0.01. What is its pH?
%%% \vspace{6mm}
I will convert the value of the lemon's hydrogen ion concentration 0.01 into a base-ten number for the logarithmic equation, which is $10^{-2}$.
%%% \vspace{6mm}
I will determine the pH by using the equation
$\mathrm{pH}=-\log[\mathrm{H}^+]$,
where the value of $[\mathrm{H}^+]$ is \qty{1e-2}{\mol\per\liter}.
\begin{align*}
\mathrm{pH}&=-\log(10^{-2})\\
&=-2\log_{10}(10)\\
&=-(-2)\\
&=2
\end{align*}
The pH of a lemon is 2.
%\vspace{12mm}
% #13
\item How does an earthquake of magnitude 8.4 compare to an earthquake of magnitude 3.7? Round the answer to the nearest whole number.
To calculate the difference in size, I will divide $10^{8.4}$ by $10^{3.7}$ since the Richter scale is logarithmic.
\begin{align*}
\mathrm{Size\_difference}&=\frac{10^{8.4}}{10^{3.7}}\\
&=10^{8.4-3.7}\\
&=10^{4.7}\\
&\approx 50119
\end{align*}
An earthquake of 8.4 magnitude on the Richter scale is approximately 50119 times more powerful than one of 3.7 magnitude on the same scale.
\end{enumerate}
\end{document}