Combined Log Law¶

Introduction:¶

Logarithms are the inverse operation to exponentials. They are useful because they change the arithmetic operations of multiplication, division, and powers into addition, subtraction, and products. While modern calculators can do all these operations quickly, for us mere humans logarithms can be useful for doing quick approximations in our heads.

Motivation:¶

Going about our day, we often run into powers of ten, when we see kilograms of food in a grocery store (1000 grams), megawatts of power from an electrical generator (1,000,000 watts) or gigabytes of memory in our computer (1,000,000,000 bytes). It is the power of ten that is important, and the logarithm captures this idea with the formulas

\[ \log(10) = 1 \]

\[ \log(1000) = 3 \]

\[ \log(1,000,000) = 6\]

\[ \log(1,000,000,000) = 9. \]

The logarithm of a number \(x\) is defined as the power \(n\) it takes so that \(x = 10^n.\) So, for instance, since \(1000 = 10^3\), we know that \(\log(1000) = 3,\) as indicated in the list above.

For numbers that aren’t integer powers of 10, the logarithm is still defined by the above formula, where \(n\) can be any real number solving \(x = 10^n\). For instance, you might guess that \(\log(5000)\) is somewhere between 3 and 4, since the number 5000 is halfway between \(10^3 = 1000\) and \(10^4 = 10,000\). You might even guess that \(\log(5000) = 3.5\), which is not a bad approximation: in fact, a calculator shows that

\[ \log(5000) = 3.69897..., \]

which is the same as saying

\[ 5000 = 10^{3.69897...}. \]

We can also take logarithms of small numbers, like this:

\[ \log(0.01) = \log\left(\frac{1}{100}\right) = \log(10^{-2}) = -2. \]

But you cannot take logarithms of negative numbers. (Unless you are willing to learn about something called complex numbers!)

Base for logarithm:¶

In the examples above, we worked with powers of ten, so ten is called the base for the logarithm.

We can work with other bases. For instance, with computers we often work with power of two. A KB of memory is actually \(1024 = 2^{10}\) bytes.

If you aren’t sure about this, multiply out \(2\times 2 \times 2 \times \ldots \times 2\) with ten 2’s, to see you get \(1024= 2^{10}.\)

A MB of memory is \(1,048,576 = 2^{20}\) bytes, or just over a million bytes. A GB is \(1073741824 = 2^{30}\) bytes, or just over a billion bytes.

It’s a funny coincidence that \(10^3 \approx 2^{10}\) so that kilo =1000 is about the same as Kilo = 1024.

We write this down in logarithm form, adding a subscript to keep track of the base. So

\[ \log_2(1024) = 10 \]

\[ \log_2(1048576) = 20 \]

\[ \log_2(1073741824) = 30. \]

In general, the number \(\log_2(x)\) is defined as the solution to

\[ x = 2^n. \]

Logarithms can be defined with any number \(B\) as a base, provided \(B\) is positive and not equal to one. The function is then written as \(\log_B(x).\)

Three important bases:¶

In practice, there are only three log functions that occur in most of math and science:

\[ \log_2(x), \log_{10}(x), \mbox{ and } \log_e(x), \]

which have bases 2, 10 and \(e\), respectively, where \(e = 2.71...\) is the natural exponential that occurs in calculus.

The base ten logarithm \(\log_{10}(x)\) occurs so often that it is sometimes abbreviated as \(\log(x)\), as we did in the first section of this notebook.

The base \(e\) logarithm is called the natural log, written \(\ln(x)\). The natural logarithm arises in calculus, where it is often denoted simply as \(\log x\). So one must pay attention to the context when the base is unspecified!

Examples:¶

{1} Can we find \(\log_2(4000)\) approximately, without using a calculator?

Sure. Here’s one way. We know that \(4 = 2^2\), and that \(1000 \approx 2^{10}\). So \(4000 \approx 2^2 \times 2^{10} = 2^{12}\). So we conclude

\[ \log_2(4000) \approx 12. \]

A quick check with a calculator shows \(\log_2(4000) = 11.96578...\) so that was a pretty good approximation!

{2} Can we find \(\log(\pi)\) approximately?

Well, our friends the ancient Egyptians thought that \(\pi\) was the square root of 10. It’s not, but that’s a pretty good approximation. So we have

\[ \log(\pi) \approx \log(10^{1/2}) = 1/2. \]

In fact, a check with a calculator shows \(\log(\pi) = 0.49715...\), so again we have a pretty good approximation.

Basics of Logarithms:
¶

Even though logarithms can seem very complicated, we can look at the basic relationship between logarithms and exponentials in order to simplify these expressions to furture enhance our understandings. Before looking deeper in these relationships, we will first identify the main components of a logarithmic function. Logarithms are written in the following form:

\(\log_B(x)=m\) where B is the base of the logarithm.

Given a number \(x\), we define \(\log_B(x)=m\) as the solution to the exponential relationship

\[ x=B^m. \]

Logarithmic Laws
¶

There are 4 main logarithmic laws which help show the relationship between exponential and logarithmic functions.

Product Law: \(\log_{B}(x \times y)=\log_{B}(x)+\log_{B}(y)\)
Quotient Law: \(\log_{B}( x \div y) =\log_{B}(x)-\log_{B}(y)\)
Power Law: \(\log_{B}(x^p)=p\times \log_B(x)\)
Changing Base Rule: \(\log_{B}(x)=\frac{\log_C(x)}{\log_C(B)}\)

Background: Exponential Laws¶

Since logarithms are closely related with exponents, we will be using exponential laws when deriving logarithmic laws. Exponential Laws state:

\(B^m \times B^n=B^{m+n} \quad (1) \)
\(\frac{B^m}{B^n}=B^{m-n} \quad \quad \; \;\;\,(2)\)
\((B^m)^n=B^{mn} \quad \quad \,(3)\)
\((BC)^m=B^m C^m \quad \,(4)\)

We will be referring to these laws throughout the program using the number in the brackets above.

from IPython.display import display, Latex, clear_output
from math import log
import matplotlib.pyplot as plt
import matplotlib.gridspec as gridspec
%matplotlib inline
import numpy as np
import ipywidgets as widgets
from ipywidgets import interact, interactive, fixed, interact_manual,IntSlider, Output, VBox, HBox, Label
from abc import ABCMeta, abstractmethod

class Logarithm:
    #############################################
    #     Description: stores all functions that will be using as shortcuts when graphing functions.
    #############################################
    
    @staticmethod
    #A static method is contains an object that does not change.
    #In this case, the functions are created to pass in arguments and should not change.
    def log(m,b,x):
        #########################################
        #   Function: y= log(mx)
        #   
        #   @Args:
        #   m: constant inside the logarithm
        #   b: base of the logarithm
        #   x: vector with all of the x-coordinates
        #
        #   @Returns:
        #   y: y-coordinates of the graph based on the x-coordinates and function
        #
        #########################################
        i=x*m
        return [log(y,b) for y in i]

    @staticmethod
    def log_exp(r,m,b,x):
        #########################################
        #   Function: y=log((mx)^r)
        #   
        #   @Args:
        #   m: constant inside the logarithm
        #   b: base of the logarithm
        #   x: vector with all of the x-coordinates
        #   r: exponent within the logarith
        #
        #   @Returns:
        #   y: y-coordinates of the graph based on the x-coordinates and function
        #
        #########################################
        x= (x*m)**r
        return [log(y,b) for y in x]

    @staticmethod
    def constant_x_log(r,m,b,x):
        #########################################
        #   Function: y=r*log(mx)
        #   
        #   @Args:
        #   r: constant multiplied by the logarithm
        #   m: constant inside the logarithm
        #   b: base of the logarithm
        #   x: vector with all of the x-coordinates
        #
        #   @Returns:
        #   y: y-coordinates of the graph based on the x-coordinates and function
        #
        #########################################
        x= x*m
        return [r*log(y,b) for y in x]

    @staticmethod
    def division_of_logs(m,n,b,x):
        #########################################
        #   Function: y=log_m(nx)/log_b(nx)
        #   
        #   @Args:
        #   m: base of logarithm in the numerator
        #   b: base of logarithm in the denominator
        #   n: constant inside each logarithm
        #   x: vector with all of the x-coordinates
        #
        #   @Returns:
        #   y: y-coordinates of the graph based on the x-coordinates and function
        #
        #########################################
        y1=Logarithm.log(m,n,x)
        y2=Logarithm.log(b,n,x)
        y=np.divide(y1,y2)
        return y

#########################################
# Variables:
# base - Value of the base of the logarithms
#      - Over the coarse of the program, we will set the base to be 10.
# x    - The range of numbers that are shown on the x-axis for the graphical proofs
#########################################

base=10
x=np.linspace(1,10)

Product Law¶

The first law we are looking at is the Product Law. This is used when finding the sum of two logarithmic functions with the same base. The law states that

\(\log_{B}(xy)=\log_{B}x+\log_{B}y\).

An example¶

\(\log(100\times 1000) = \log(100) + \log(1000)\) or equivalently
\(\log(100,000) = 5 = 2 + 3.\)

Mathematical proof¶

We will look at the mathematical proof. It may look complicated, however, it can simply be broken down. First we fix quantities \(x,y\) and then define

\(p=\log_B x\) and \(q=\log_B y\).

The equivalent exponential forms are

\(B^p=x\) and \(B^q=y\).

We take the product of these two equations to obtain

\(B^p \times B^q=x \times y\),

and from the Exponential Law (1), we can get the equivalent expression

\(B^{p+q}=x \times y\).

We apply log to both sides

\(\log_B(B^{p+q})=\log_B(x \times y),\)

and then by the definition of a logarithm, we have

\(p+q=\log_B(x \times y)\).

Since we know \( p=\log_B x\) and \( q=\log_B y\), we obtain

\(\log_{B}x+\log_{B}y = \log_{B}(x \times y).\)

That completes the mathematical proof of the product law.

Graphical Demonstration¶

As we know, the product law states: \(\log_{B}x+\log_{B}y = \log_{B}(x \times y).\)

To go about this, we introduce a parameter \(t\) that allows us to trace the graph of the logarithm function. We also introduce two constant integers, \(m\) and \(n\).

We let \(x=mt\) and \(y=nt\) and set the base \(B\) to 10, abbreviating \(\log_{10}(x)\) as \(\log(x)\).

For these values of \(x\) and \(y\), the product law becomes \(\log(mt)+\log(nt) = \log(mnt^2)\)

For the graphical demonstration, we graph the three terms in the above equation separately with respect to \(t\). When looking at a \(t\) value, the sum of the corresponding values of the functions on the left side of the equation should be equivalent to the function on the right side of the equation, thus providing a demonstration of the Product Law.

class ProductLaw():
    # Create 2x2 sub plots
    gs = gridspec.GridSpec(2, 2)
    
    axis=5
    x=6
    y=3
    
    x_axis_bar = widgets.IntSlider(
        value=5,
        min=1,
        max=10,
        step=1,
        description='$t$',
        disabled=False,
        continuous_update=False,
        orientation='horizontal',
        readout=True,
        readout_format='d'
    )       

    x_bar = widgets.IntSlider(
        value=x,
        min=1,
        max=10,
        step=1,
        description='$m$',
        disabled=False,
        continuous_update=False,
        orientation='horizontal',
        readout=True,
        readout_format='d'
    )            

    y_bar = widgets.IntSlider(
        value=y,
        min=1,
        max=10,
        step=1,
        description='$n$',
        disabled=False,
        continuous_update=False,
        orientation='horizontal',
        readout=True,
        readout_format='d'
    )   
    
    def create_graph():

        #########################################
        # Description: generates a graph in order to prove Product Law
        #
        # @Args: Inputs are for the variables shown in ProductLaw
        #        x-coordinate: based on a sliding bar in range 1-10
        #        M constant: based on a sliding bar in range 1-10
        #        N constant: based on a sliding bar in range 1-10
        #
        # @Return: graph for graphical proof as well as the y-coordinate corresponding to the graphed points
        #
        #########################################
        
        #Plot the 3 log functions from the left and right side of the Product Law
        ax1 = plt.subplot(ProductLaw.gs[0, 0]) # row 0, col 0
        ax1.plot(x,Logarithm.log(ProductLaw.x,base,x),'-b',label='$y=\log_{B}(x)$')
        p1=log(ProductLaw.x*ProductLaw.axis,base)
        ax1.plot(ProductLaw.axis,p1,'ob')
        ax1.annotate('%1.3f' %p1,xy=(ProductLaw.axis,p1),xytext=(-10,-20),textcoords='offset points')
        ax1.set_title('Left side of Product law')
        plt.ylabel('$\log_{B}(mt)$')
        ax1.yaxis.set_label_position("left")
        plt.grid()

        ax2 = plt.subplot(ProductLaw.gs[1, 0])
        ax2.plot(x,Logarithm.log(ProductLaw.y,base,x),'-g',label='$y=\log_{B}(y)$')
        p2=log(ProductLaw.y*ProductLaw.axis,base)
        ax2.plot(ProductLaw.axis,p2,'og')
        ax2.annotate('%1.3f' %p2,xy=(ProductLaw.axis,p2),xytext=(-10,-20),textcoords='offset points')
        plt.ylabel('$\log_{B}(nt)$')
        ax2.yaxis.set_label_position("left")
        plt.xlabel('$t$')
        plt.grid()

        ax3 = plt.subplot(ProductLaw.gs[:, 1])
        ax3.plot(x,Logarithm.log(ProductLaw.x*ProductLaw.y,base,x**2),'-r',label='$y=\log_{B}(xy)$')
        p3=log(ProductLaw.x*ProductLaw.y*(ProductLaw.axis**2),base)
        ax3.plot(ProductLaw.axis,p3,'or')
        ax3.annotate('%1.3f' %p3,xy=(ProductLaw.axis,p3),xytext=(-10,-20),textcoords='offset points')
        ax3.set_title('Right side of Product Law')
        plt.ylabel('$\log_{B}(mnt^2)$')
        ax3.yaxis.set_label_position("right")
        plt.xlabel('$t$')
        plt.grid()
        plt.show()
    
        display(Latex('When $m$={1:1d} and $n$={2:1d}'.format(ProductLaw.axis,ProductLaw.x,ProductLaw.y)))

        #Display the value of the points to prove that the law is valid
        display(Latex('From the marked y-coordinates on the graph above, the points at log({0:1d}$t$), log({1:1d}$t$) and log({2:1d}$t^2$) are at {3:1.3f}, {4:1.3f} and {5:1.3f} repectively'.format(ProductLaw.x,ProductLaw.y,ProductLaw.x*ProductLaw.y,p1, p2, p3)))
        display(Latex('{0:1.3f}+{1:1.3f}={2:1.3f}'.format(p1,p2,p1+p2)))
        display(Latex('{0:1.3f}={1:1.3f}'.format(p3,p3)))
        display(Latex('This means that the left side of the equation equals the right side'))
        display(Latex('thus'))
        display(Latex(r'$\log_{B}x+\log_{B}y = \log_{B}(x \times y)$'))

    def clear_display():  
        clear_output(wait=True)
        display(ProductLaw.x_bar)
        display(ProductLaw.y_bar)
        display(ProductLaw.x_axis_bar)
        ProductLaw.create_graph()
        ProductLaw.observe()
        
    def observe():
        ProductLaw.x_axis_bar.observe(ProductLaw.xv, names='value')
        ProductLaw.x_bar.observe(ProductLaw.x_barv, names='value')
        ProductLaw.y_bar.observe(ProductLaw.y_barv, names='value')
        #ProductLaw.clear_display()
        
    def xv(value):
        ProductLaw.axis=value['new']
        ProductLaw.clear_display()
        
    def x_barv(value):
        ProductLaw.x=value['new']
        ProductLaw.clear_display()
        
    def y_barv(value):
        ProductLaw.y=value['new']
        ProductLaw.clear_display() 

ProductLaw.clear_display()

../../../_images/combined-log-law_15_3.png

\[\text{When } m=6\text{ and }n=3\]

From the marked y-coordinates on the graph above, the points at \(\log(6t), \log(3t)\) and \(\log(18t^2)\) are at 1.477, 1.176 and 2.653 repectively.

\[1.477+1.176=2.653\]

\[2.653=2.653\]

This means that the left side of the equation equals the right side. Thus

\[\log_{B}x+\log_{B}y = \log_{B}(x \times y)\]

Results¶

In the mathematical proof, we used the relationship between logarithms and exponents in order to derive the Product Law. Based on the values recorded during the graphical proof, we see that the left-hand side of the law is equivalent to sum of the two functions on the right-hand side.

Callysto Curriculum Notebooks

Logarithmic Laws

Combined Log Law¶

Introduction:¶

Motivation:¶

Base for logarithm:¶

Three important bases:¶

Examples:¶

Basics of Logarithms:
¶

Logarithmic Laws
¶

Background: Exponential Laws¶

Product Law¶

An example¶

Mathematical proof¶

Graphical Demonstration¶

Results¶

Quotient Law¶

An example¶

Mathematical proof¶

Graphical Demonstration¶

Result¶

Power Law¶

An example¶

Mathematical Proof¶

Graphical Demonstration¶

Results¶

Change of Base Rule¶

An example¶

Mathematical Proof¶

Discussion¶

Evidence¶

Results¶

Examples¶

1. Simplify the following equation, then solve for \(x\)¶

\( 3^{\log(x)}3^{\log(x)}\)¶

2. \(\text{Simplify the expression: } 2\log(x) - \frac{\log(z)}{2} + 3\log(y)\)¶

3. Solve for x \( 2^{(x-2)}=2^x -2 \)¶

Conclusion¶

Callysto Curriculum Notebooks

Logarithmic Laws

Combined Log Law¶

Introduction:¶

Motivation:¶

Base for logarithm:¶

Three important bases:¶

Examples:¶

Basics of Logarithms:¶

Logarithmic Laws¶

Background: Exponential Laws¶

Product Law¶

An example¶

Mathematical proof¶

Graphical Demonstration¶

Results¶

Quotient Law¶

An example¶

Mathematical proof¶

Graphical Demonstration¶

Result¶

Power Law¶

An example¶

Mathematical Proof¶

Graphical Demonstration¶

Results¶

Change of Base Rule¶

An example¶

Mathematical Proof¶

Discussion¶

Evidence¶

Results¶

Examples¶

1. Simplify the following equation, then solve for \(x\)¶

\( 3^{\log(x)}3^{\log(x)}\)¶

2. \(\text{Simplify the expression: } 2\log(x) - \frac{\log(z)}{2} + 3\log(y)\)¶

3. Solve for x \( 2^{(x-2)}=2^x -2 \)¶

Conclusion¶

Basics of Logarithms:
¶

Logarithmic Laws
¶