Change of Base and Evaluation
Key Definitions
- Change of Base Formula: For any positive bases $a$, $b$ ($a \neq 1$, $b \neq 1$) and $x > 0$, $$ \log_a x = \frac{\log_b x}{\log_b a} $$ This formula allows conversion of logarithms from one base to another.
- Base Conversion Identity: For $a > 0$, $a \neq 1$, $b > 0$, $b \neq 1$, $$ \log_a b = \frac{1}{\log_b a} $$ This is a special case of the change of base formula.
- Chain Rule of Logarithms: For $a, b, c > 0$ and $a, b \neq 1$, $$ \log_a b \cdot \log_b c = \log_a c $$
- Base Power Conversion: For $a > 0$, $a \neq 1$, $b > 0$, and $k \neq 0$, $$ \log_{a^k} b = \frac{1}{k} \log_a b $$
- Argument Power Conversion: For $a > 0$, $a \neq 1$, $b > 0$, and $k \neq 0$, $$ \log_a (b^k) = k \log_a b $$
Theory and Concepts
The Change of Base Formula
The change of base formula is one of the most powerful tools in logarithmic manipulation. It allows us to evaluate logarithms in any base using calculators (which typically have only $\log_{10}$ and $\ln$ functions) and to simplify complex logarithmic expressions.
Derivation:
Let $\log_a x = y$. Then by definition, $a^y = x$.
Taking logarithm base $b$ on both sides:
$$ \log_b (a^y) = \log_b x $$ $$ y \log_b a = \log_b x $$ $$ y = \frac{\log_b x}{\log_b a} $$Hence, $\log_a x = \dfrac{\log_b x}{\log_b a}$.
Important Consequences
- $\log_a b = \dfrac{1}{\log_b a}$ (by taking $x = b$ in the change of base formula)
- $\log_{a^m} b^n = \dfrac{n}{m} \log_a b$ (by applying base and argument power conversions)
- For any three positive numbers $a, b, c$ ($a, b \neq 1$): $$ \log_a b \cdot \log_b c \cdot \log_c a = 1 $$
- $\log_a x = \dfrac{\ln x}{\ln a} = \dfrac{\log x}{\log a}$ (using natural or common logarithm)
Evaluation Strategy
To evaluate logarithmic expressions efficiently:
- Convert all logarithms to a common base (preferably base 10 or base $e$)
- Apply power rules to simplify arguments
- Use the identity $\log_a a = 1$ to reduce expressions
- Combine terms using product and quotient rules
Visual Representation of Base Conversion
Solved Examples
Solved Example 3.1
Evaluate $\log_8 32$ using change of base formula.
Solution:
Method 1 (Using base 2):
$$ \log_8 32 = \frac{\log_2 32}{\log_2 8} = \frac{\log_2 2^5}{\log_2 2^3} = \frac{5}{3} $$Method 2 (Using base 10):
$$ \log_8 32 = \frac{\log 32}{\log 8} = \frac{\log 2^5}{\log 2^3} = \frac{5\log 2}{3\log 2} = \frac{5}{3} $$Solved Example 3.2
Simplify: $\frac{1}{\log_2 36} + \frac{1}{\log_3 36}$.
Solution:
Using $\frac{1}{\log_a b} = \log_b a$:
$$ \frac{1}{\log_2 36} + \frac{1}{\log_3 36} = \log_{36} 2 + \log_{36} 3 $$ $$ = \log_{36} (2 \times 3) = \log_{36} 6 $$Now, $36 = 6^2$, so:
$$ \log_{36} 6 = \log_{6^2} 6 = \frac{1}{2} \log_6 6 = \frac{1}{2} $$Solved Example 3.3
If $\log_{12} 27 = a$, find $\log_6 16$ in terms of $a$.
Solution:
Given $a = \log_{12} 27 = \frac{\log 27}{\log 12} = \frac{3\log 3}{\log 3 + 2\log 2}$
Let $x = \log 2$, $y = \log 3$. Then:
$$ a = \frac{3y}{y + 2x} $$We need $\log_6 16 = \frac{\log 16}{\log 6} = \frac{4\log 2}{\log 2 + \log 3} = \frac{4x}{x + y}$
From $a = \frac{3y}{y + 2x}$, we get $a(y + 2x) = 3y \Rightarrow ay + 2ax = 3y \Rightarrow 2ax = (3 – a)y \Rightarrow \frac{y}{x} = \frac{2a}{3 – a}$
Therefore:
$$ \log_6 16 = \frac{4x}{x + y} = \frac{4}{1 + \frac{y}{x}} = \frac{4}{1 + \frac{2a}{3 – a}} = \frac{4}{\frac{3 – a + 2a}{3 – a}} = \frac{4(3 – a)}{3 + a} $$Solved Example 3.4
Find the value of $\log_3 4 \cdot \log_4 5 \cdot \log_5 6 \cdot \log_6 7 \cdot \log_7 8 \cdot \log_8 9$.
Solution:
Using the chain rule $\log_a b \cdot \log_b c = \log_a c$ repeatedly:
$$ \begin{align*} &\log_3 4 \cdot \log_4 5 = \log_3 5\\ &\log_3 5 \cdot \log_5 6 = \log_3 6\\ &\log_3 6 \cdot \log_6 7 = \log_3 7\\ &\log_3 7 \cdot \log_7 8 = \log_3 8\\ &\log_3 8 \cdot \log_8 9 = \log_3 9 = 2 \end{align*} $$Solved Example 3.5
If $\log_a b = 2$, $\log_b c = 3$, $\log_c d = 4$, find $\log_d a$.
Solution:
We have:
$$ \log_a b \cdot \log_b c \cdot \log_c d \cdot \log_d a = 1 $$Substituting the given values:
$$ 2 \times 3 \times 4 \times \log_d a = 1 $$ $$ 24 \log_d a = 1 \Rightarrow \log_d a = \frac{1}{24} $$Solved Example 3.6
Evaluate: $\log_2 3 \cdot \log_3 4 \cdot \log_4 5 \cdots \log_{15} 16$.
Solution:
Using the chain rule:
$$ \log_2 3 \cdot \log_3 4 = \log_2 4 = 2 $$ $$ \log_4 5 \cdot \log_5 6 = \log_4 6 $$ $$ \log_6 7 \cdot \log_7 8 = \log_6 8 $$ $$ \log_8 9 \cdot \log_9 10 = \log_8 10 $$ $$ \log_{10} 11 \cdot \log_{11} 12 = \log_{10} 12 $$ $$ \log_{12} 13 \cdot \log_{13} 14 = \log_{12} 14 $$ $$ \log_{14} 15 \cdot \log_{15} 16 = \log_{14} 16 $$The product becomes:
$$ 2 \cdot \log_4 6 \cdot \log_6 8 \cdot \log_8 10 \cdot \log_{10} 12 \cdot \log_{12} 14 \cdot \log_{14} 16 $$Now $\log_4 6 \cdot \log_6 8 = \log_4 8 = \frac{3}{2}$
$$ \log_8 10 \cdot \log_{10} 12 = \log_8 12 $$ $$ \log_{12} 14 \cdot \log_{14} 16 = \log_{12} 16 $$So we have: $2 \times \frac{3}{2} \times \log_8 12 \cdot \log_{12} 16 = 3 \times \log_8 16 = 3 \times \frac{4}{3} = 4
Solved Example 3.7
If $\log_{10} 2 = a$ and $\log_{10} 3 = b$, express $\log_5 6$ in terms of $a$ and $b$.
Solution:
$$ \log_5 6 = \frac{\log_{10} 6}{\log_{10} 5} = \frac{\log_{10} (2 \times 3)}{\log_{10} \left(\frac{10}{2}\right)} $$ $$ = \frac{\log_{10} 2 + \log_{10} 3}{\log_{10} 10 – \log_{10} 2} = \frac{a + b}{1 – a} $$Solved Example 3.8
Solve: $\log_x 2 \cdot \log_{x/16} 2 = \log_{x/64} 2$.
Solution:
Using change of base to base 2:
$$ \log_x 2 = \frac{1}{\log_2 x}, \quad \log_{x/16} 2 = \frac{1}{\log_2 (x/16)} = \frac{1}{\log_2 x – 4} $$ $$ \log_{x/64} 2 = \frac{1}{\log_2 (x/64)} = \frac{1}{\log_2 x – 6} $$The equation becomes:
$$ \frac{1}{\log_2 x} \cdot \frac{1}{\log_2 x – 4} = \frac{1}{\log_2 x – 6} $$Let $t = \log_2 x$:
$$ \frac{1}{t(t-4)} = \frac{1}{t-6} \Rightarrow t-6 = t(t-4) $$ $$ t-6 = t^2 – 4t \Rightarrow 0 = t^2 – 5t + 6 $$ $$ t^2 – 5t + 6 = 0 \Rightarrow (t-2)(t-3) = 0 \Rightarrow t = 2 \text{ or } t = 3 $$Therefore, $x = 2^2 = 4$ or $x = 2^3 = 8$.
Solved Example 3.9
Prove that $\log_a x = \frac{\log_b x}{\log_b a}$ and hence evaluate $\log_{27} 81$.
Solution:
The formula is proven in the theory section. Using it:
$$ \log_{27} 81 = \frac{\log_3 81}{\log_3 27} = \frac{\log_3 3^4}{\log_3 3^3} = \frac{4}{3} $$Solved Example 3.10
If $a, b, c$ are positive real numbers not equal to 1, and $\log_a b + \log_b a = 2$, find the relation between $a$ and $b$.
Solution:
Let $\log_a b = t$. Then $\log_b a = \frac{1}{t}$.
Given $t + \frac{1}{t} = 2 \Rightarrow t^2 + 1 = 2t \Rightarrow t^2 – 2t + 1 = 0 \Rightarrow (t-1)^2 = 0 \Rightarrow t = 1$.
Therefore, $\log_a b = 1 \Rightarrow b = a$.
Concept Application Exercise 3.1
- Evaluate: $\log_{16} 128$.
- Simplify: $\frac{1}{\log_3 5} + \frac{1}{\log_7 5} – \frac{1}{\log_{21} 5}$.
- If $\log_{12} 27 = a$, find $\log_6 16$ in terms of $a$.
- Find the value of $\log_2 3 \cdot \log_3 4 \cdot \log_4 5 \cdots \log_{15} 16$.
- If $\log_a b = 3$ and $\log_b c = 2$, find $\log_c a$.
- Given $\log_{10} 2 = 0.3010$ and $\log_{10} 3 = 0.4771$, evaluate $\log_5 6$.
- Solve: $\log_x 3 \cdot \log_{x/9} 3 = \log_{x/27} 3$.
- If $\log_{30} 3 = a$ and $\log_{30} 5 = b$, express $\log_{15} 8$ in terms of $a$ and $b$.
- Prove that $\log_a x \cdot \log_b y = \log_b x \cdot \log_a y$ for $x, y > 0$.
- Find the value of $\log_{0.1} 0.001 + \log_{0.01} 0.0001$.