Logarithms
Matsec Syllabus: Pure Mathematics Paper 1 Topic 1: Logarithms:
Definition of logarithms;
The laws of logarithms;
Common and natural logarithms;
Change of base formula;
Solution of equations involving indices and logarithms.
Introduction
The following is the definition of the logarithm function:
The logarithm (or log, for short) of a number is a quantity representing the power (index/exponent) to which the base must be raised to produce that given number.
This definition means that an equation involving indices can be converted into an equation involving logs as follows:
ax=b⟺loga(b)=x where b is a positive number.
Notes:
- The symbol ⟺ means that the two equations (in this case, ax=b and loga(b)=x) are equivalent (so they describe the same thing).
- The number a of the logarithmic function loga(b) (written as a subscript) is known as the base of the logarithmic function.
Example 1: Convert the following equations into ones involving logs:
(a) 102=100
Substitute a=10, x=2, b=100 in the definition.
(b) 105=10000
Substitute a=10, x=5, b=10000 in the definition.
(c) 43=64
Substitute a=4, x=3, b=64 in the definition.
(d) e4.605=100
Substitute a=e, x=4.605, b=100 in the definition.
Example 2: Convert the following equations into ones involving indices:
(a) log381=4
Substitute a=3, x=4, b=81 in the definition.
(b) log1010000=4
Substitute a=10, x=4, b=10000 in the definition.
(c) log42=21
Substitute a=4, x=21, b=2 in the definition.
(b) loge1=0
Substitute a=e, x=0, b=1 in the definition.
The Common & Natural Log
The logarithmic function with base 10, namely log10, is known as the common logarithm and is written as log in short.
The logarithmic function with base e, namely loge, is known as the natural logarithm and is written as ln in short. The letter “n” in “ln”, stands for natural.
Example 3: Convert the following equations into ones involving indices:
(a) log100=2
log100log10100=2=2
Substitute a=10, x=2, b=100 in the definition.
(b) lne=1
lnelogee=1=1
Substitute a=e, x=1, b=e in the definition.
Example 4: Evaluate the following without using a calculator:
(a) log464
Let log464=x
We want to find the value of x.
(b) log162
Let log162=x
We want to find the value of x.
(c) logaa3
Let logaa3=x
We want to find the value of x.
(d) log100000
Let log10100000=x
We want to find the value of x.
Two Results of Logs
We have the following 2 results of logs that help us evaluate logs, more quickly without referring back to the indices.
Result 1 states that whenever the input of the log function is equal to its base, the output is always equal to 1.
Result 2 states that whenever the input of the log function is equal to 1, the output is always equal to 0.
Result 1:
logaa=1 for any positive number a
Result 2:
loga1=0 for any positive number a
We are going to use these 2 results in order to evaluate logs.
Example 5: Evaluate the following without using a calculator:
(a) log55
(b) 20log88
(c) lne
(d) log10
(e) log1
(f) ln1
(e) alog41
(f) 21ln1
Three Rules of Logs
The following 3 rules of logs are very useful in simplifying and evaluating expressions involving logs.
Rule 1 states that the log of a product of two terms can be expressed as the sum of the log of the two terms.
Similarly, Rule 2 states that the log of a fraction of two terms can be expressed as the log of the numerator minus the log of the denominator.
Rule 3, states that when the input of the log function is in index form, the power can be removed from the input of the log function and added as a coeeficient of the log function.
Rule 1:
logcab=logca+logcb where a, b & c are positive numbers
Rule 2:
logcba=logca−logcb where a, b & c are positive numbers
Rule 3:
logcab=blogca where a, b & c are positive numbers
Let us now use these three rules in order to evaluate expressions involving logs.
Example 6: Simplify the following expressions:
(a) log3a+log35
(b) log4x−logx
(c) logx−1
(d) log10−log3+log60
(e) log210−log26
(f) 2log4−log2+3
(g) 4log32−2
(h) 2lna−ln2a
Example 7: Evaluate the following expressions:
(a) log464
(b) log364log316
Avoid the following common mistakes
Avoid the following common mistakes related to the use of the 3 Rules of Logs:
∙∙∙∙∙ logc(a+b)=logca+logcb logc(a−b)=logca−logcb logc(ab)=(logca)(logcb) logca2=(logca)2 logc(ba)=logcblogca→→→→→ The correct rule is Rule 1: logc(ab)=logca+logcb The correct rule is Rule 2: logc(ba)=logca−logcb The correct rule is Rule 1: logc(ab)=logca+logcb Using Rule 3, the correct equation is: logca2=2logca The correct rule is Rule 2: logc(ba)=logca−logcb
Change of Base Formula
The following is the formula that enables us to change the base of the log function. Here we are changing the base from b to c.
logba=logcblogca
For example, suppose that we have log35 and we would like continue the working in base 4. Then by the Change of Base Formula:
log35=log43log45
Note that the Change of Base Formula was very handy in times when the calculator only evaluated log base 10 functions. Thus the Change of Base Formula was used mostly to convert a log of any base to base 10. Nowadays, our calculators and computers evaluate log functions of any base straight away, and thus this formula is mostly used to simply expressions and solving equations.
Example 7: Express the following with logs base 10 and evaluate:
(a) log749
(b) log11121
(c) log2439
(d) log422
(e) (log25)(log58)
(f) logab2logba3
Example 8: Use the Change of Base formula to show that: logba=logab1
logba=logablogaa (by Change of Base formula)=logab1 (by Result 1)
Solving indical equations by using logs
In the section on Indices we solved indical equations, that is, equations in which the unknown variable is in the power. For example, 2x=16. We used the fact that 16=24. Hence we obtained both the left hand side and the right-hand side as an index having the same base. See worked example below.
Sometimes we can have an indical equation in which this is not possible, for example when we have 2x=15 and thus we solve it by using logs. The trick is to put a log on each side of the equation and Rule 3, in order to bring down x from the power. See the worked examples, in Example 9.
2x2xx=16=24=4
2x=15
Here we cannot express 15 in index form with base 2. Hence we must use logs.
Example 9: Solve the following equations:
(a) 2x=15
(b) 3x+1=4
(c) 2x=31−x
* R.H.S. stands for right-hand-side (of the equation)