New Zealand
Level 8 - NCEA Level 3

# Applications of Logarithmic Functions

Lesson

The main feature of logarithmic growth is that it gets slower and slower. There are many processes in biology, engineering, finance, and elsewhere that can be described by logarithmic functions.

Often it will be helpful to consider transformations of the equation $y=\log_bx$y=logbx so that the resulting equation better describes the process being modelled. In this lesson we will look at equations of the form $y=a\log_b\left(x-h\right)+k$y=alogb(xh)+k, which involve vertical dilation, horizontal translation, and vertical translation.

#### Exploration

The Trinket and Widget Corporation (TWC) is exploring ways to increase the productivity of its workforce. Initial research shows that the number of widgets, $N$N, manufactured by the average TWC warehouse each month is related to the average monthly employee salary, $S$S, by the equation $N_1=40\log_2\left(S+1\right)$N1=40log2(S+1).

 Graph of $N_1=40\log_2\left(S+1\right)$N1​=40log2​(S+1).

The graph of the equation passes through the origin, which suggests that the employees are not willing to build any widgets if they have no salary. Notice that as the average monthly salary increases, the number of widgets produced also increases, but this number grows less and less with every increase in salary.

In an effort to produce more widgets, the management team at TWC send out an employee survey to find ways they can improve the working conditions. They find that providing free fruit for the employee lunchroom leads to a slight increase in productivity, which is modelled by the equation $N_2=60\log_2\left(S+1\right)$N2=60log2(S+1).

Let's compare the two equations on the same set of axes. The graph of the equation $N_2=60\log_2\left(S+1\right)$N2=60log2(S+1) corresponds to a stretching in the vertical direction of the graph of $N_1=40\log_2\left(S+1\right)$N1=40log2(S+1).

 Graph of $N_1=40\log_2\left(S+1\right)$N1​=40log2​(S+1) and graph of $N_2=60\log_2\left(S+1\right)$N2​=60log2​(S+1).

The management team can only go so far with free fruit. They want to consider the long term future of their workforce and have decided to trial a new fleet of robots to work on the widget assembly line.

The robots don't need a salary, and can reliably produce a large quantity of the most technically advanced components of the widget each month. The new productivity relationship is described by the equation $N_3=60\log_2\left(S+16\right)$N3=60log2(S+16).

On the axes below we show the graphs of the equation $N_2=60\log_2\left(S+1\right)$N2=60log2(S+1) and the equation $N_3=60\log_2\left(S+16\right)$N3=60log2(S+16). We can see that the graph describing the productivity with the robots is a horizontal translation of the graph describing the productivity resulting from only the free lunchroom fruit.

 Graph of $N_2=60\log_2\left(S+1\right)$N2​=60log2​(S+1) and graph of $N_3=60\log_2\left(S+16\right)$N3​=60log2​(S+16).

In general, a horizontal translation of the graph of $y=\log_bx$y=logbx is represented in the form $y=\log_b\left(x-h\right)$y=logb(xh), where the resulting graph is translated $h$h units to the right, or by the form $y=\log_b\left(x+h\right)$y=logb(x+h), where the graph is translated $h$h units to the left.

In the example above, the graph of $N_2=60\log_2\left(S+1\right)$N2=60log2(S+1) is translated by $15$15 units to the left to produce the graph of $N_3=60\log_2\left(S+16\right)$N3=60log2(S+16).

Now suppose that new industry standards require the TWC to perform quality assurance tests on their widgets. Each month the engineering department will take $50$50 widgets at random off the production line and run through strength and durability assessments. These widgets will not be able to be sold to consumers, so this will decrease the output from the warehouse.

The productivity relationship that accounts for the quality assurance testing is described by the equation $N_4=60\log_2\left(S+16\right)-50$N4=60log2(S+16)50. The graph of $N_3$N3 and the graph of $N_4$N4 are shown below.

 Graph of $N_3=60\log_2\left(S+16\right)$N3​=60log2​(S+16) and graph of $N_4=60\log_2\left(S+16\right)-50$N4​=60log2​(S+16)−50.

Notice that the graph of $N_4=60\log_2\left(S+16\right)-50$N4=60log2(S+16)50 is the result of a vertical translation of the graph of $N_3=60\log_2\left(S+16\right)$N3=60log2(S+16) by $50$50 units downward.

#### Worked example

Following the productivity research, the Trinket and Widget Corporation has decided to implement all the recommendations of the management team. The widget productivity and employee salary relationship is now modelled by the equation $N=60\log_2\left(S+16\right)-50$N=60log2(S+16)50.

• How many widgets will be produced if no salary is paid to employees?

If no salary is paid, then the value of S will be zero. When we substitute this into the equation we have  $N$N $=$= $60\log_216-50$60log2​16−50 $=$= $60\times4-50$60×4−50 $=$= $240-50$240−50 $=$= $190$190
So there will still be $190$190 widgets produced each month.
• If the average salary is $\$500$$500 per month, what is the expected number of widgets produced? By substituting S=500S=500 into the equation we have  NN == 60\log_2\left(500+16\right)-5060log2​(500+16)−50 == 60\log_2516-5060log2​516−50 == 491491 (to nearest whole widget) The warehouse will produce around 490490 widgets if the employees are given \500$$500 per month.
• How many more widgets will be produced if the monthly salary is doubled to $\$1000$$1000? First we can find the number of widgets produced when the salary is \1000$$1000, then we can subtract from this the $491$491 widgets expected when the salary is only $\$500$$500.  NN == 60\log_2\left(1000+16\right)-5060log2​(1000+16)−50 == 60\log_21016-5060log2​1016−50 == 549549 (to nearest whole widget) By doubling the salary, the warehouse will produce and extra 549-491=58549491=58 widgets. #### Practice question A major communications company found that the more they spend on advertising, the higher their revenue. Their sales revenue, in thousands of dollars, is given by R=10+20\log_4\left(x+1\right)R=10+20log4(x+1), where xx represents the amount they spend on advertising (in thousands of dollars). 1. Determine their sales revenue if they spend no money on advertising. 2. Determine their sales revenue if they spend \14000$$14000 on advertising.

Give your answer to the nearest thousand $dollars$dollars.

3. Would you say that every extra $\$1000$$1000 spent on advertising becomes more or less effective in terms of raising revenue? more effective: sales revenue increases as advertising spending increases. A less effective: every extra \1000$$1000 spent on advertising raises the sales revenue by less and less

B

more effective: sales revenue increases as advertising spending increases.

A

less effective: every extra $\$10001000 spent on advertising raises the sales revenue by less and less

B

### Outcomes

#### M8-4

Use curve fitting, log modelling, and linear programming techniques

#### M8-7

Form and use trigonometric, polynomial, and other non-linear equations

#### 91574

Apply linear programming methods in solving problems

#### 91575

Apply trigonometric methods in solving problems