LOG GRAPHS

LOGARITHMS (LOGS)

Intro

Index Form and Log Form

• Index form: 100 = 10² : where 100 = the number, 10= the base & 2 = the power (exponent)

• Log form: log₁₀ 100 = 2 : where 100 = the number, 10= the base & 2 = the power (exponent)

• Other examples:

• Index : 49 = 7² Log : log₇ 49 = 2

• Index : 27 = 3³ Log : log₃ 27 = 3

Laws of Logs (5 Log Laws )

1. Product
  $Log\ AB\ =\ Log\ A\ +\ Log\ B\$  

Eg: Log 5 + Log 4 = Log 20 
 $5\ \times4\ =20$  

2. Quotient 
 $Log\ \frac{A}{B}=\ \log\ A\ -\ Log\ B\$  
Eg: Log 28 - Log 4 = Log 7 
 $\frac{28}{4}=7$  

3.Power
 $\log\ x\ ^n=n\ \log\ x$  

Eg :
 $\log\ 5^2\ =2\ \log\ 5\$  

4.Base = Number 
 $\log_aa\ =1$  
Eg:  $\log_33\ =\ 1$  

5.  Power = 0 
 $\log_a\ 1\ =\ 0$  
Eg:  $\log_{5\ }1\ =\ 0$  

Linear Reduction / Log Graphs (2 Types)

 Equations such as  $y=\ Kx^{n\ }\ \&\ \ y\ =\ Ab^x$  are exponential equations, not straight lines.  However an equation of this type can be converted using logarithms to a linear form.

y = mx + c 

Type 1

 $y\ =\ kx^n\ \left(non-linear\right)$  
(Take Logs)  $Log\ y\ =\ \log\ kx^n$  
                       $Log\ y\ =\ \log\ k\ +\ \log\ x^n$  
                       $Log\ y\ =\ \log\ x^{n\ }+\ \log\ k$   

              $\left(linear\right)Log\ y\ =\ n\ \log\ x\ +\ \log\ k\$      

                                     y   =   m   x            +    c                                

1. In this type, both the x and y values given are logged.

2. In this type, k and n are values  to be determined. (If they aren't already given.)

A plotted graph helps to determine this.

Type 2

(non- linear)  $y\ =Ab^{x\ }\$  

(Take Logs )    $Log\ y\ =\ Log\ Ab^x$  
                        $Log\ y\ =\ Log\ A\ +\ Log\ b^x$  
                       $Log\ y\ =\ Log\ b^{x\ }+\ Log\ A$  
                      $Log\ y\ =\ x\ Log\ b\ +\ Log\ A\$ 
                     $Log\ y\ =\ Log\ b\ \left(x\right)\ +Log\ A$  (linear)
                              y    =    m         x     +    c
1. In this type only the y values are logged because the "x" in the linear equation is not in log form.
2. In this type, the A and b is to be determined. Sometimes the x value is given.
 A plotted graph helps to determine this. 

Outline of the graph.

(Note to self, zoom in to be clearly seen)

Example of how a Type 1 Log Graph Equation is worked.

 Question:
a) Convert the equation to linear form.  $y\ =kx^n$  

b) Using the values of log x and log y, plot a suitable graph and hence determine the values of x and n.

Steps

1.Using the x and y values table the question gives, find the log x and log y values to 2d.p. using a calculator.

2.Convert the non linear equation to a linear equation by taking logs. (See slide 6)

3.Using the log x and log y values, plot the points on a graph sheet and draw the best fit straight line cutting the log y-axis.

 4.Find K which is determined based on the value on the log y-axis that the straight line cuts. (in this example it was 0.165)

 $\therefore\log_{10}\ K\ =\ 0.165$  

 $\therefore K\ =\ 10^{0.165}$  
     $K\ =\ 1.46\ 2d.p$  

5.Find n. n is the gradient of the straight line on the graph. The formula is  $n=\frac{y_2\ -y_1}{x_2-x_1}$  

  $n=\frac{0.44-0.25}{0.50-0.16}$  

 $n=\frac{0.19}{0.34}$  

 $n=0.56$  

Answer

 6.Now that k and n was determined, the answer is written as :

Ans =  $y=kx^n$  

       = $1.46x^{0.56}$   

Example of how a Type 2 Log Graph Equation is worked.

Question:

 $Ab^x$  
a) Express this equation in a suitable form that can produce a straight line graph.

b) Use the graph to determine the value of A and b.

c) Find the value of y when x = 0.8.

Steps

1.Using the x and y values table the question gives, find the log y values to 2d.p. using a calculator. (Remember this type only the y values are logged )

2.Convert the non linear equation to a linear equation by taking logs. (See slide 7 )

3.Using the x and log y values, plot the points on a graph sheet and draw the best fit straight line cutting the log y-axis.

4.Find A which is determined based on the value on the log y-axis that the straight line cuts. (in this example it was 0.90) 

 $\therefore Log_{10}\ A\ =\ 0.90$  

  $\therefore A\ =\ 10^{0.90}$  

      $A=7.94\ 2d.p$  

 5.Find log b. log b is the gradient of the straight line on the graph. The formula is   $\frac{y_2\ -y_1}{x_2-x_1}$ 

 

  $\log\ b\ =\ \frac{0.8-0.4}{0.5-2.05}$  

   $\log\ b\ =\ \frac{0.4}{-1.55}$  

 $b\ =\ -0.26$  

Answer

 6.Now that A and b is determined the answer can be written as:

Ans = $y\ =\ Ab^x$  

        $y\ =\ 7.94\left(-0.26\right)^{0.8}$  
             $y\ =\ 2.70\ 2d.p$