Logarithmic and logarithmic functions and models
This tutorial: Part B: Logarithmic functions and models
(This topic is also in Section 2.4 in Finite Mathematics and Applied Calculus)
Resources
Logarithmic functions: Basics
Logarithmic function
Logarithmic functions have the following form:
$f(x) = \log_b x + C$ \t \gap[10] $C$ %%and $b$ #[are arbitrary constants with $b$ positive and not equal to 1.][son constantes arbitrarias con $b$ positivo y no igual a 1.]#
\\ #[Technology formula][Fórmula tecnológica]#: \t \gap[10] log(x)/log(b)+C %%or ln(x)/ln(b)+C
#[Alternative forms][Formas alternativas]#
We can use the logarithm identities from the %%logstut write $\log_b x$ in the form
\t $\dfrac{\log x}{\log b} = \dfrac{1}{\log b}\log x = A\log x$ \t $A = \dfrac{1}{\log b} ={}$ #[constant][constante]#
\\ #[or][o]# \t $\dfrac{\ln x}{\ln b} = \dfrac{1}{\ln b}\ln x = A\ln x$ \t $A = \dfrac{1}{\ln b} ={}$ #[constant][constante]#
giving us two alternative forms of a general logarithmic function
$f(x) = A\log x + C$
\\ $f(x) = A\ln x + C$ \t $\ln x = \log_e x$ is the natural logarithm. See %%ExpFuncsB for more on the number $e$ and the natural logarithm.
where $A$ and $C$ are arbitrary constants with $A \ne 0$.
Their graphs have the form shown below. When $b \gt 1$ $f(x)$ increases with increasing $x$, and when $b \lt 1$ it decreases with increasing $x$:
$\bold{f(x) = \log_b x}$ \t
\\ 

Adding a constant $C$ has the effect of shifting the graphs vertically $C$ units:
$f(x) = \log_b x$ #[Increasing when ][Aumentando cuando ]# $b \gt 1$
\t $f(x) = \log_b x$ #[Decreasing when ][Disminuyendo cuando ]# $b \lt 1$
$\bold{f(x) = \log_b x + C}$ \t
\\ 

#[Some features seen in the above graphs][Algunas características que se ve en las gráficas anteriores]#
$f(x) = \log_b x + C$ #[Increasing when ][Aumentando cuando ]# $b \gt 1$
\t $f(x) = \log_b x + C$ #[Decreasing when ][Disminuyendo cuando ]# $b \lt 1$
- As $\log_b 1 = 0$, the graph of $\log_b x$ crosses the $x$-axis at $x = 1.$
- If $b \gt 1$, the values $x = b, b^2, b^3, ...$ form an increasing sequence and the corresponding values of $\log_b x$ are
$\log_b(b) = 1$ \\ $\log_b(b^2) = 2\log_b(b) = 2$ \\ $\log_b(b^3) = 3\log_b(b) = 3$ \\ ... \\ $\log_b(b^n) = n\log_b(b) = n.$
- If $b \lt 1$, the values $x = b^{-1}, b^{-2}, b^{-3}, ...$ form an increasing sequence and the corresponding values of $\log_b x$ are
$\log_b(b^{-1}) = -\log_b(b) = -1$ \\ $\log_b(b^{-2}) = -2\log_b(b) = -2$ \\ $\log_b(b^{-3}) = -3\log_b(b) = -3$ \\ ... \\ $\log_b(b^{-n}) = -n\log_b(b) = -n.$
%%Examples
1. #[The Logarithmic function][La función logarítmica]# $f(x) = \log_2 x - 1$ #[has][tiene]# $b = 2$ %%and $C = -1.$ #[as $b \gt 1$ the values of $f$ increase with increasing $x.$][ya que $b \gt 1$ los valores de $f$ aumentan al aumentar $x.$]# #[Here is a table calculating various values of $f$ and the resulting graph:][Aquí hay una tabla que calcula varios valores de $f$ y la gráfica resultante:]#
2. #[One for you][Uno para ti]#
#[Application: Earthquakes][Aplicación: Temblores]#
#[The magnitude of a large earthquake can be measured by the formula][La magnitud de un gran terremoto se puede medir con la fórmula]#
$\displaystyle M = \frac{2}{3}\log S - 10.7,$ \gap[10] \t Moment magnitude seismic scale
where $S$ is the seismic moment, which measures the intensity of the earthquake based on the slippage and size of the fault and the rigidity of the fault material.†
† $S$ is measured in ergs, which are units also used to measure energy. (An erg is the amount of energy it takes to accelerate a stationary mass of 2 grams to a velocity of 1 cm/sec.)
(a) Calculate the seismic moment of a 6.0 magnitude earthquake. (b) The earthquake in Lice, Turkey on July 28 1976 had a magnitude of 6.0 and the earthquake in Kerman, Iran on December 26 2003 had a magnitude of 6.6. Compare the two: The seismic moment in the Turkey earthquake was what percentage of the seismic moment of the Iran quake? #[Solution][Solutión]# (a)
$M = \dfrac{2}{3}\log S - 10.7$
\\ $6.0 = \dfrac{2}{3}\log S - 10.7$
We need to solve this equation for $M_0$.:
$\dfrac{2}{3}\log S = 6.0 + 10.7 = 16.7$ \gap[20] \t Move the 10.7 over.
\\ $\log S = \dfrac{3}{2}(16.7) = 25.05$ \gap[20] \t Multiply by $3/2$.
\\ $S = 10^{25.05} \approx 1.12 \times 10^{25}$ #[ergs][ergios]#. \gap[20] \t Exponent form
(b) #[Take $S_1$ to be the seismic moment in the Iran quake and $S_2$ to be the moment in the Turkey quake. The calculation we just did can be written as][Toma $S_1$ como el momento sísmico del terremoto de Irán y $S_2$ como el momento del terremoto de Turquía. El cálculo que acabamos de hacer se puede escribir como]#
$M_1 = 10^{(3/2)(\color{indianred}{6.0} + 10.75)}$ \t #[Turkey][Turquía]#
\\ $M_2 = 10^{(3/2)(\color{indianred}{6.6} + 10.75)}$ \t #[Iran][Irán]#
\\ $\dfrac{S_1}{S_2} = 10^{(3/2)(\color{indianred}{6.0} + 10.75) -(3/2)(\color{indianred}{6.6} + 10.75)} \qquad$ \t #[Laws of exponents][Leyes de los exponentes]#
\\ \gap[20] ${}= 10^{(3/2)(\color{indianred}{6.0 - 6.6})}$ \t #[Simplify][Simplifica]#
\\ \gap[20] ${}= 10^{(3/2)(-0.6)} = 10^{-0.9} \approx 0.13$
#[Thus, the seismic moment in the Turkey quake was about 13% of that in the Iran quake.][Por lo tanto, el momento sísmico en el terremoto de Turquía fue aproximadamente el 13% del del terremoto de Irán.]#
Now try some of the exercises in Section 2.4 in Finite Mathematics and Applied Calculus.
or move ahead by pressing "Next tutorial" on the sidebar.
Copyright © 2021 Stefan Waner and Steven R. Costenoble