Applications of logs I

We might be curious about the relationship between population and time, or the number of configurations of a number plate and its length. These relationships are either exponential or logarithmic, depending on which quantity we choose as the subject of the relationship.

From a description of these relationships, we want to be able to identify its logarithmic equation and use the equation to draw a graph, or to infer quantities in the relationship.

Exploration

• Consider a number plate that has $n$n characters where each character can be any one of the $26$26 letters or the $10$10 digits. How can we construct an equation that expresses $n$n in terms of the number of possible combinations, $x$x?
  
  Well, in each position there are $36$36 possible characters available. Consider a number plate with $7$7 characters, like the one shown below. There are $36\times36\times36\times36\times36\times36\times36$36×36×36×36×36×36×36 possible combinations, which we can write more simply as $36^7$367 combinations.

  $7$7 digit number plate

  
  In general, we can say that there are $36^n$36n total possible combinations for a number plate of length $n$n, and so we have the equation

  $x=36^n$x=36n

  We can make $n$n the subject by rewriting the equation in logarithmic form:

  $n=\log_{36}x$n=log36​x

  
  This equation may be valuable when forecasting where we might need to increase the number of characters for a growing population.
   
• How many characters would be needed for a country with a population of $2$2 million?
  
  A quick substitution will tell us the number of characters needed for $2$2 million different possible number plates:

  $n$n	$=$=	$\log_{36}x$log36​x	(Writing down the equation)
  $n$n	$=$=	$\log_{36}2000000$log36​2000000	(Substituting)
  $n$n	$\approx$≈	$4.0487...$4.0487...	(Evaluating)
  $n$n	$=$=	$5$5	(Rounding up)

  Since we can only have a whole number of characters, we want to round our number up. This will account for more than $2$2 million possible number plates - if we had rounded down instead, there wouldn't be enough unique number plates for everyone in the population.
  
  Now consider the graph of the logarithm function given below.

  Graph of $n=\log_{36}x$n=log36​x

• If the population doubles every year, how many years will it take until they have to increase the number of characters used?
  
  Well after one year, the population will be $4$4 million, then it will become $8$8 million and so on. One strategy is to first use a point on the graph to identify when $n$n first exceeds $5$5. Well we can see that graph roughly crosses $n=5$n=5 when $x=60$x=60 million.

  Point indicated at $x=60$x=60 million

  
  If we substitute $x=60$x=60 million into the equation to check, we get a value of $n=4.9978\dots$n=4.9978… which is very close to our value of $n=5$n=5, but falls a little short.
  Since the population increases in powers of $2$2, we know that after $6$6 years, the population will be $2^6=64$26=64 million. Substituting $x=64$x=64 million into the equation we get

  $n$n	$=$=	$\log_{36}64000000$log36​64000000	(Substituting)
  $n$n	$\approx$≈	$5.015\dots$5.015…	(Simplifying)
  $n$n	$=$=	$6$6	(Rounding up)

  
  So after $6$6 years, the population will increase to $x=64$x=64 million. At this point, licence plates that are only $5$5 digits long will not be enough to account for everyone in the population, and so the country will have to start issuing $6$6 digit licence plates.

Practice question