Logarithms are the inverse of exponents. They are used to determine the power to which a base must be raised to obtain a given number.
Mathematical definition:
$\boldsymbol{\log_{b}a = c \iff b^{c} = a}$
Where:
Take Note :
1. b > 0 (The base must be positive)
2. b ≠ 1 (The base cannot be 1)
3. a > 0 (The number/argument must be positive)
In simple terms, a logarithm answers the question: "To what power must the base be raised to get a?"
(i) Exponential Form.
This form expresses a number as a base raised to a power.
$\boldsymbol{b^{c} = a}$
(ii) Logarithmic Form.
This form expresses the exponent required to produce a number.
$\boldsymbol{\log_{b}a = c}$
(iii) Common Logarithm.
This is a logarithm with base 10.
$\boldsymbol{\log x = \log_{10}x}$
(iv) Natural Logarithm.
This is a logarithm with base e.
$\boldsymbol{\ln x = \log_{e}x}$
The value of e is approximately 2.718.
The following laws are used to simplify logarithmic expressions. All laws apply only when the arguments are positive and the base is greater than zero but not equal to 1.
(i) Product Law.
$\boldsymbol{\log_{b}(MN) = \log_{b}M + \log_{b}N}$
This law states that the logarithm of a product is equal to the sum of the logarithms of the individual factors.
(ii) Quotient Law.
$\boldsymbol{\log_{b}\left(\frac{M}{N}\right) = \log_{b}M - \log_{b}N}$
This law states that the logarithm of a quotient is equal to the difference of the logarithms.
(iii) Power Law.
$\boldsymbol{\log_{b}(M^{k}) = k \log_{b}M}$
This law states that the logarithm of a number raised to a power is equal to the exponent multiplied by the logarithm of the base number.
(iv) Change of Base Law.
$\boldsymbol{\log_{b}a = \frac{\log a}{\log b}}$
This law allows logarithms to be rewritten in another base, usually base 10 or base e, for easier evaluation.
(v) Special Logarithmic Values.
$\boldsymbol{\log_{b}1 = 0}$, because $\boldsymbol{b^{0} = 1}$.
$\boldsymbol{\log_{b}b = 1}$, because $\boldsymbol{b^{1} = b}$.
(i) Solving Logarithmic Equations.
To solve logarithmic equations, convert the logarithmic form into exponential form.
Example:
$\boldsymbol{\log_{2}x = 3}$
$\boldsymbol{2^{3} = x}$
$\boldsymbol{x = 8}$
(ii) Equating Logarithms with the Same Base.
If $\boldsymbol{\log_{b}M = \log_{b}N}$, then $\boldsymbol{M = N}$.
This is valid only when the base is the same and the arguments are positive.
(iii) Combining Logarithms.
$\boldsymbol{\log a + \log b = \log(ab)}$
$\boldsymbol{\log a - \log b = \log\left(\frac{a}{b}\right)}$
These rules are derived from the product and quotient laws.
(iv) Expanding Logarithms.
$\boldsymbol{\log(ab^{2}) = \log a + 2\log b}$
This process breaks down complex logarithmic expressions into simpler parts.
(v) Domain of Logarithmic Functions.
The argument of a logarithm must always be positive.
Therefore:
Example 1
Solve:
$\boldsymbol{\log_{2}x = 4}$
Step 1: Convert to exponential form.
$\boldsymbol{2^{4} = x}$
This step changes the logarithmic equation into an exponential equation.
Step 2: Evaluate the power.
$\boldsymbol{x = 16}$
Raising 2 to the power of 4 gives the value of x.
Example 2
Solve:
$\boldsymbol{\log_{3}(x) = \log_{3}(9)}$
Step 1: Remove the logarithms.
$\boldsymbol{x = 9}$
When logarithms have the same base, their arguments are equal.
Example 3
Solve:
$\boldsymbol{\log x + \log 2 = 1}$
Step 1: Apply the product law.
$\boldsymbol{\log(2x) = 1}$
The sum of logarithms becomes the logarithm of a product.
Step 2: Convert to exponential form.
$\boldsymbol{10^{1} = 2x}$
Base 10 is used since no base is written.
Step 3: Simplify.
$\boldsymbol{10 = 2x}$
Evaluating the power gives a simple equation.
Step 4: Solve for x.
$\boldsymbol{2x = 10}$
$\boldsymbol{x = 5}$
Example 4
Solve:
$\boldsymbol{\log_{2}(x + 1) = 3}$
Step 1: Convert to exponential form.
$\boldsymbol{2^{3} = x + 1}$
This removes the logarithm and simplifies the equation.
Step 2: Evaluate the power.
$\boldsymbol{8 = x + 1}$
Raising 2 to the power of 3 gives 8.
Step 3: Solve for x.
$\boldsymbol{x = 7}$
Subtract 1 from both sides to obtain the value of x.
Example 5
Solve:
$\boldsymbol{\log(x - 1) + log(x - 3) = 1}$
Step 1: Apply the product law.
$\boldsymbol{\log[(x - 1)(x - 3)] = 1}$
The sum of logarithms becomes the logarithm of a product.
Step 2: Convert to exponential form.
$\boldsymbol{10^{1} = (x - 1)(x - 3)}$
The logarithmic equation is converted into exponential form.
Step 3: Simplify.
$\boldsymbol{10 = (x - 1)(x - 3)}$
Evaluating the power gives a quadratic equation.
Step 4: Expand the brackets.
$\boldsymbol{10 = x^{2} - 4x + 3}$
This removes the brackets and simplifies the expression.
Step 5: Rearrange the equation.
$\boldsymbol{x^{2} - 4x - 7 = 0}$
All terms are brought to one side of the equation.
Step 6: Solve using the quadratic formula.
$\boldsymbol{x = \frac{4 \pm \sqrt{44}}{2}}$
The quadratic formula is used to find the values of x.
Step 7: Simplify.
$\boldsymbol{x = 2 \pm \sqrt{11}}$
The solution is simplified to its final form.
Attempt the following Questions:
1. Convert to exponential form:
$\boldsymbol{\log_{3} 81 = 4}$
2. Expand completely:
$\boldsymbol{\log(2x^{2}y)}$
3. Simplify:
$\boldsymbol{\log_{10}100 + \log_{10}10}$
4. Solve:
$\boldsymbol{\log_{2}x = 5}$
5. Simplify:
$\boldsymbol{\log(50) - \log(5)}$
6. Convert to logarithmic form:
$\boldsymbol{7^{2} = 49}$
7. Solve:
$\boldsymbol{\log_{3}x = \log_{3}9}$
8. Expand completely:
$\boldsymbol{\log\left(\frac{4x^{3}}{y}\right)}$
9. Evaluate:
$\boldsymbol{\log_{5}1}$
10. Use change of base to evaluate:
$\boldsymbol{\log_{2}8}$
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