Tutorial: Logarithms
Game version
(This topic is also in Section 0.8 in Finite Mathematics and Applied Calculus)

Logarithms are exponents
Recall from the %%exponentstut that, in the expression $b^n$, $b$ is called the base, and $n$ is called the exponent. Thus, when, for example, we write

$2^3 = 8$,

The exponent to which you must raise the base 2 in order to obtain 8 is 3.

The logarithm base 2 of 8 is 3.
$\log_2 8 = 3 \qquad$  #[The logarithm with base 2][El logarítmo en base 2]#  #[of 8][de 8]#  #[equals 3][es igual a 3]#. 
↑  ↑  ↑  
#[The exponent to which you must raise the base 2][El exponente al que debes elevar el base 2]#  #[to obtain 8][para obtener 8]#  #[equals 3][es igual a 3]#. 
Base b logarithm
The base $b$ logarithm of $x$, $\log_bx$, is the exponent to which you need to raise $b$ in order to obtain $x$. In symbols,
$\log_bx = y$ \t #[means][significa]# \t $b^y=x$.
\\ Logarithm form \t \t Exponential form
%%Note
1. $\log_b x$ is only defined if $b$ %%and $x$ are both positive, and $b \neq 1$.
2. $\log_{10} x$ is called the common logarithm of $x$ and is sometimes written as $\log x$ Suggested video for this topic: Video by HCCMathHelp
1. $\log_b x$ is only defined if $b$ %%and $x$ are both positive, and $b \neq 1$.
2. $\log_{10} x$ is called the common logarithm of $x$ and is sometimes written as $\log x$ Suggested video for this topic: Video by HCCMathHelp
%%Examples
The following table lists some exponential equations and their equivalent logarithmic form.
Exponential form  Logarithm form 
$4^2=16$  $\log_416=2$ 
$2^3=8$  $\log_28=3$ 
$10^3=1{,}000$  $\log_{10}1{,}000=3\ $ %%or $\ \log 1{,}000 = 3$ 
$5^1=5$  $\log_55=1$ 
$7^0=1$  $\log_71=0$ 
$4^{2}=\dfrac{1}{16}$  $\log_4\left(\dfrac{1}{16}\right)=2$ 
$10^{1}=\dfrac{1}{10}$  $\log_{10}\left(\dfrac{1}{10}\right)=1\ $ %%or $\ \log\left(\dfrac{1}{10}\right)=1$ 
$25^{1/2}=5$  $\log_{25}5=\dfrac{1}{2}$ 
Algebra of logarithms
The laws of exponents in the %%exponentstut can be translated into rules for manipulating logarithms:
Logarithm identities
#[The following identities work for all positive bases $a \neq 1$ and $b \neq 1$, all positive numbers $x$ and $y$, and every real number $r$.][Las siguientes identidades son válidas para todas las bases positivas $a \neq 1$ y $b \neq 1$, todos los números positivos $x$ y $y$, y cada número real $r$.]#
Suggested video for this topic: Video by Don't Memorise
Identity  Examples 
$\log_b(xy)$${}= \log_bx + \log_by$
The logarithm of a product equals the sum of the logarithms. 

$\log_b\left(\dfrac{x}{y}\right) = \log_bx  \log_by$
#[The logarithm of a quotient equals the difference of the logarithms.][El logaritmo de un cociente es la resta de los logaritmos.]# 

$\log_b(x^r) = r\log_bx$
The logarithm of an $r$th power is $r$ times the logarithm. 

$\log_bb = 1$ #[and][y]# $\log_b1 = 0$
#[The base $b$ logarithm of $b$ is $1$, and the logarithm of $1$ is $0$.][El logaritmo de $b$ en base $b$ es $1$, y el logaritmo de $1$ es $0$.]# 

$\log_b\left(\dfrac{1}{x}\right) = \log_bx$
The logarithm of a reciprocal is the negative of the logarithm. 

$\log_bx = \dfrac{\log_ax}{\log_ab}$
"Change of base:" The base $b$ logarithm of $x$ is the logarithm of $x$ over the logarithm of the base. 

Solving for unknowns in the exponent
Logarithms are very useful in solving equations where the unknown is in the exponent. First go through the following example and both methods of solution, and then try the others on your own:
%%Example:
#[Solve for $x$:][Despeja a $x$:]# $\quad 5^{2x} = \dfrac{1}{125}$.
#[Solution Method 1: Take the logarithm of both sides (always works).][Solución méodo 1: Toma el logarítmo de ambos lados (siempre funciona).]#
The given equation \gap[20] \t $5^{2x} = \dfrac{1}{125}$
\\ Take the logarithm of both sides. \gap[20] \t $\log (3^{2x})=\log \left(\dfrac{1}{125}\right)$
\\ Use the algebra of logarithms to simplify. \gap[20] \t $2x\log 5 = \log 125$
\\ Solve for $x$. \gap[20] \t $x=\dfrac{\log 125}{2\log 5}$
In this case, we can again use the algebra of logarithms to get the answer in simpler form:
$x=\dfrac{\log 125}{2\log 5}$ \t ${}= \dfrac{\log (5^3)}{2\log 5}$
\\ \t ${}= \dfrac{3\log 5}{2\log 5}$ \t \gap[40] #[Logarithm of a power][Logaritmo de una potencia]#
\\ \t ${}= \dfrac{3}{2}$ \t \gap[40] #[Cancel the][Cancela el]# $\log 5$.
#[Solution Method 2: Convert to logarithmic form (works only if the the equation can be written in the form $b^c = A$, but leads to a simplerlooking answer more rapidly than Method 1).][Solución méodo 2: Convertir a forma logarítmico (funciona solo si se puede escribir la ecuación en la forma $b^c = A$, pero lleva a una respuesta simple más rápido que el método 1).]#
The given equation. \gap[20] \t $5^{2x} = \dfrac{1}{125}$
\\ Make sure it is written in the form $b^c = A$. \gap[20] \t Yes it is! \\ Rewrite it in logarithmic form. \gap[20] \t $2x=\log_5\left(\dfrac{1}{125}\right)$
\\ Simplify and/or evaluate the logarithm.* \gap[20] \t $2x=\log_5 125 = \log_5(5^3) = 3$
\\ Solve for $x$. \gap[20] \t $x=\dfrac{3}{2}$
* If evaluating the logarithm results in an irrational number, it it often better to leave it as a logarithm rather than use a decimal approximation.
Now try the exercises in Section 0.8 in Finite Mathematics and Applied Calculus.
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Copyright © 2020 Stefan Waner and Steven R. Costenoble