Basic Algebra
3.9 Exponential and Logarithmic functions
Examples
Here are some examples of sigma notation
1. 4
X
k=0
2k = 20 + 21+ 22 + 23+ 24 = 31
2. ∞
X
k=4
k = 4 + 5 + 6 + 7 + 8 + · · · 3.
P5
k=2(−1)k+1 31k
= (−1)3 312 + (−1)4 313 + (−1)5 314 + (−1)6 315
= − 312 + 313 − 314 + 315
3.8.3 Factorials
The notation n!, where n is an integer greater or equal to 0, is spoken “n factorial” and is a shorthand for
n × (n − 1) × (n − 2) × · · · × 2 × 1.
So for example
5! = 5 × 4 × 3 × 2 × 1 = 120.
It should be clear that this time of expressions gets very large, very quickly, and indeed 69! is so large that most calculators are unable to repre-sent it.
We note that by convention we accept 1! = 1 and 0! = 1.
3.9 Exponential and Logarithmic functions
We now consider another, slightly more complex type of function.
3.9.1 Exponential functions
In quadratics, we had x terms with specific constant powers, such as x2. A very powerful function is formed when x is in the exponent (yet another name for “power” or “index”).
An exponential function is one of the form
3.9 Exponential and Logarithmic functions 33
y = kx
where k is some positive constant. These functions are important due to their extraordinary ability to climb or decrease, as we shall see later when we examine their graphs.
3.9.2 Logarithmic functions
For a positive number n, the logarithm or log of n to base b, written logbn is the power to which b must be raised to give n. To put this in mathematics:
logbn = x ⇔ bx = n
The antilogarithm or antilog of n to the base b is bn. There are two main “types” of logarithms in use today:
• log10, often written simply as log;
• loge, often written1 as ln.
Examples
Logs are often used to condense a very large range of numbers to a more managable one.
Examine how in the following table the logs of a large range of values from one thousandth, to one thousand are contracted to a range from -3 to 3. Verifying that there logarithms are correct is left as a simple exercise to the reader.
Value 0.001 0.01 0.1 1 10 100 1000
Log10 -3 -2 -1 0 1 2 3
The scales of pH (chemical scale of acidity) and the decibel range, the Richter scale of earthquake intensity are all logarithmic,2 base 10.
Therefore a pH of 6 is 10 times more acidic than the neutral 7, and 5 is 100 times more acidic than 7, etc.
1The Scottish mathematician John Napier (1550 - 1617) did a great deal of work on methods of computation and published his Mirifici logarithmorum canonis descripto (Discription of the marvellous rule of logarithms) in 1614. In his honour logs base e are often called Naperian logarithms. Logs were first used to help perform large calculations.
2A lesser known example is the warpspeed scale
3.9 Exponential and Logarithmic functions 34
Warnings
Just as we have problems with the square roots of negative numbers, so we cannot take the logarithms of negative numbers, or of zero.
Laws of Logarithms
Logarithms are so useful because they exhibit the following properties, known as the laws of logarithms. These are closely related to the laws of indices and are shown in table 3.4.
1 logn(x × y) = logn(x) + logn(y) 2 logn(x ÷ y) = logn(x) − logn(y) 3 logn(xy) = y logn(x)
4 lognn = 1
5 logn1 = 0
Table 3.4: Laws of Logarithms
These laws easily account for their use in computation. To multiply two large numbers, one would find their logarithms from tables, add these two values, and then by the first law, this is the logarithm of the product, so one merely antilogs this number to obtain the answer.
The other major application ot these is in solving equations where the variable is in the index, i.e. some sort of exponential expression.
3.9.3 Logarithms to solve equations
When we obtain equations like this ax = b
it is sometimes possible to guess the power to which we must raise a to obtain b. However, more often than not, a and b are not simple integers, and therefore almost impossible to guess x. In this case, take logarithms on both sides of the equation.
3.9 Exponential and Logarithmic functions 35
⇒ lognax = lognb
⇒ x logna = lognb (using laws of logs 3)
⇒ x = loglognb
na
3.9.4 Examples
Solve the following equations for x.
1.
3x= 27 2.
3x= 30 3.
22x− 5(2x) + 6 = 0
Solutions
1. Although it is clear here that the answer is x = 3 we use logs to illustrate the point. It really doesn’t matter what base of logs we use, provided we are consistent. Take logs both sides
log 3x= log 27 ⇒ x log 3 = log 27
using the third law of logs (see 3.9.2). We now can perform ordinary rearranging.
x = log 27 log 3 = 3
2. This is only slightly more complicated, although it is not possible to guess the answer easily.
log 3x= log 30 ⇒ x log 3 = log 30 x = log 30
log 3 = 3.0959 to four decimal places.
3. This problem looks very complicated. Taking logs immediately will get us into trouble. Partly because we have no law to split logs over addition (note this carefully, a common mistake), and that the log of zero is a problem (it’s something like −∞).
We might observe that the equation roughly resembles a quadratic (see 3.7, and this is the key. Let us assign u = 2x, then clearly
3.9 Exponential and Logarithmic functions 36
u = 2x ⇒ u2 = (2x)2 = 22x
by the laws of indices (see 3.5). Placing these into the equation yields u2 − 5u + 6 = 0
which is now a plain quadratic. Solving by whatever method yields u = 2 and u = 3 as solutions. Having dealt with the quadratic, we now deal with the exponential problem, recall that u = 2x.
Consider the u = 2 solution, this means 2x = 2 so that clearly x = 1, with no logarithms required.
The u = 3 solution proceeds as follows 2x = 3
and following the procedure above we obtain x = log 3
log 2 = 1.5850
to four decimal places. So we have two solutions x = 1, x = 1.5850.
3.9.5 Anti-logging
Logarithms are used to undo expressions in which the x we wish to obtain is in the power. Similarly, exponential functions may be used to undo logarithms.
Recall our definition.
logbn = x ⇔ bx = n this means that
blogbn = n
To put it another way, we can remove a logarithm by taking the base number to the power of the logarithm. Let’s use some numbers to illustrate the point.
Example We know that
log10(1000) = −3
3.9 Exponential and Logarithmic functions 37
so suppose we were asked to solve the equation log10x = −3
we know that x = 1000, but we might not notice this in a more difficult problem, or one we had not previously seen. To remove the log, we take the base number (in this case we are using base 10) and raise it to both sides:
10log10x = 10−3.
Now we know that the log and the anti-log (the process of raising the base to this power) cancel each other out on the left, so we obtain
x = 10−3 = 1000 and our equation is solved.
3.9.6 Examples
Solve the following equations for x:
1.
log x2 = 2 2.
4 ln x = 70
Solutions
1. In the absence of a specific base, we assume base 10, as noted above. So we may antilog both sides directly here, raising 10 to both sides.
10log x2 = 102 On the LHS, the anti-log and log cancel, leaving
x2 = 102 = 100 ⇒ x = ±10
2. In this case, the base of the logarithm is e. We could immediately take anti-logs on both sides, but the 4 in front of the ln x makes it messy. It is easier to rearrange to ln x first (onion within onion as before).
ln x = 70 4