Logarithms

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\title{Logarithms}
\author{Rithwik Palivela (hotstuffFTW)}

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\section{Introduction}

Logarithms are a shortcut for exponents, which are a shortcut for multiplication (so technically logarithms are a shortcut for multiplication). Logarithms are expressed as $\log_ab$; $a$ is the \textbf{base} of the logarithm and $b$ is the \textbf{argument} of the logarithm. \textbf{The base and argument of a logarithm have to be positive}. There are two special cases of logarithms; when $a = 10$, we write the logarithm as $\log b$ and when $a = e$, we write the logarithm as $\ln b$ (for more information on the number $e$, visit the link at the end of the article). Say we have the equation $a^x = b$. We can rewrite this as $x = \log_ab$. What this means is when $a$ is raised to the $x$th power, the result is $b$. In this article, we will learn how to solve basic logarithmic equations, prove many properties of logarithms, and apply our knowledge to some problems.

\section{Solving Basic Logarithmic Equation}

We'll start our study of logarithms by solving a basic logarithmic equation to get a feel for logarithms.

Solve the equation $\log_232 = x$.

We can rewrite the equation as $2^x = 32$. Since $2^5 = 32$, $\boxed{x = 5}$.

\section{Logarithmic Properties}

In this section, we will prove the many logarithmic properties. One of the main things you should learn from the following proofs is the extreme importance of the simple tool of setting $\log_ab = x$.

\subsection{The Addition Property}

Prove that $\log_ab + \log_ac = \log_a{bc}$.

Let's tackle the left-hand side term-by-term. Let $\log_ab = x \Longrightarrow a^x = b$ and let $\log_ac = y \Longrightarrow a^y = c$. Multiplying these two equations gives $a^{x + y} = bc$. Taking $\log_a$ to both sides, we get $x + y = \log_a{bc}$. However, since $x = \log_ab$ and $y = \log_ac$, we have proved that $\log_ab + \log_ac = \log_a{bc}$.

\subsection{The Subtraction Property}

Prove that $\log_ab - \log_ac = \log_a{\frac{b}{c}}$.

We proved the last property by setting $\log_ab$ and $\log_ac$ each to a different variable. We try to use this strategy again. Let $\log_ab = x \Longrightarrow a^x = b$ and $\log_ac = y \Longrightarrow a^y = c$. Dividing the first equation by the second equation, we get $\frac{a^x}{a^y} = \frac{b}{c} \Longrightarrow a^{x - y} = \frac{b}{c}$. Taking $\log_a$ to both sides, we get $x - y = \log_a\frac{b}{c}$. However, since $x = \log_ab$ and $y = \log_ac$, we have proved that $\log_ab - \log_ac = \log_a\frac{b}{c}$.

\subsection{The Argument-Power Property}

Prove that $\log_ab^n = n\log_ab$.

Once again, we let $\log_ab = x \Longrightarrow a^x = b$. Raising both sides to the $n$th power, we have that $a^{xn} = b^n$. Taking $\log_a$ to both sides, we get $xn = \log_a{b^n}$. However, since $x = \log_ab$, we have proved that $n\log_ab = \log_a{b^n}$.

\subsection{The Inverse Property}

Prove that $\log_ab = \frac{1}{\log_ba}$.

Again, we let $\log_ab = x \Longrightarrow a^x = b$. Taking $\log_b$ to both sides, we get $\log_ba^x = 1 \Longrightarrow x\log_ba = 1 \Longrightarrow x = \frac{1}{\log_ba}$. However, since $x = \log_ab$, we have proved that $\log_ab = \frac{1}{\log_ba}$.

\subsection{The Power Property}

Prove that $\log_{a^n}b^n = \log_ab$.

This time, we let $\log_{a^n}b^n = x \Longrightarrow a^{xn} = b^n$. Taking the $n$th root of both sides we get $a^x = b$. We can rewrite this as $\log_ab$, so we have proved that $\log_{a^n}b^n = \log_ab$.

\subsection{The Change-of-Base Formula}

Prove that $\frac{\log_cb}{\log_ca} = \log_ab$.

We let $\log_ab = x \Longrightarrow a^x = b$. Taking $\log_c$ to both sides, we have that $x \log_ca = \log_cb \Longrightarrow x = \frac{\log_cb}{\log_ca}$. However, since $x = \log_ab$, we have proved that $\log_ab = \frac{\log_cb}{\log_ca}$.

\subsection{Exercises}

\begin{enumerate}
\item Prove that $(\log_ab)(\log_bc) = \log_ac$ (this is called the \textbf{chain rule}).
\item Prove that $(\log_ab)(\log_cd) = (\log_ad)(\log_cb)$.
\item If I only have a calculator that evaluates base-10 logarithms, is it possible for me to evaluate $\log_23$? If it is possible, how so?
\end{enumerate}

\section{Applications}

\subsection{Simple Equations}
\begin{itemize}
\item Find all $x$ such that $\log_2(x + 2) + \log_2x = 3$.
\end{itemize}

We can simplify the equation to $\log_2x(x + 2) = 3 \Longrightarrow x(x + 2) = 8$. Solving the quadratic, we get the two solutions for $x$ as $-4$ and $2$. However, since $x$ has to be positive, $-4$ is extraneous, so our only solution is $\boxed{x = 2}$.

\begin{itemize}
\item Simplify $\log_43 + \log_46 - \log_49$.
\end{itemize}

We can rewrite the equation as $\log_4\left(\frac{3 \times 6}{9}\right) \Longrightarrow \log_42 = \boxed{\frac{1}{2}}$.

\subsection{Exercises}

\begin{enumerate}
\item Find all $x$ such that $\log 6x - \log (x + 3) = \log (x + 1)$.
\item What is the value of $(\log_23)(\log_34)(\log_45)(\log_56)\cdots(\log_{62}63)(\log_{63}64)$?
\end{enumerate}

\subsection{Advanced Equations}

\begin{itemize}
\item Find all the solutions to the equation $x^{\log x} = \frac{x^3}{100}$ (source: AHSME 1962).
\end{itemize}

Taking $\log_x$ to both sides results in $\log x = \log_x{\frac{x^3}{100}}$. We can rewrite the right-hand side as $\log_xx^3 - \log_x100$, which simplifies further to $3 - 2\log_x10$. Our equation becomes $\log x = 3 - 2\log_x10$. Remembering the Inverse Property, we rewrite $\log_x10$ as $\frac{1}{\log x}$. We now have $\log x = 3 - \frac{2}{\log x}$. We can set $y = \log x$, so we have $y = 3 - \frac{2}{y} \Longrightarrow y^2 - 3y + 2 = 0$, so $y = 2$ and $y = 1$. These two values of $y$ result in $x$-values of $100$ and $10$, respectively. Our answers are $\boxed{10}$ and $\boxed{100}$. 

\begin{itemize}
\item For all positive numbers $x \ne 1$, simplify $\frac{1}{\log_3x} + \frac{1}{\log_4x} + \frac{1}{\log_5x}$ (source: AHSME 1978).
\end{itemize}

The Inverse Property helps us rewrite the equation as $\log_x3 + \log_x4 + \log_x5$. We can further simplify this to $\log_x(3 \cdot 4 \cdot 5) = \boxed{\log_x60}$.

\subsection{Exercises}

\begin{enumerate}
\item Suppose that $p$ and $q$ are positive numbers for which $\log_9p = \log_{12}q = \log_{16}(p + q)$. What is the value of $\frac{q}{p}$ (source: AHSME 1988)?
\item If $60^a = 3$ and $60^b = 5$, then find $12^{[(1-a-b)/(2(1-b)]}$ (source: AHSME 1983).
\item If $\log 36 = a$ and $\log 125 = b$, express $\log \frac{1}{12}$ in terms of $a$ and $b$ (source: MA$\theta$ 1992).
\end{enumerate}

\section{Closing Thoughts}

I hope this guide to logarithms and their applications was helpful! If you want to learn more about logarithms, check out some of the following links. 

\section{Links for Additional Information}

\begin{itemize}
\item http://aops-cdn.artofproblemsolving.com/products/aops-vol2/exc1.pdf
\item http://artofproblemsolving.com/wiki/index.php/E
\item http://artofproblemsolving.com/wiki/index.php/Logarithm
\end{itemize}

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