Change of Base in Logarithms: Formulas with Examples
The proof of the change of base formula for logarithms is presented below, along with clear examples showing its application in evaluating and rewriting mathematical expressions. The formula is essential whenever you need to compute a logarithm whose base is not available on your calculator.
Table of Contents
Prerequisites and Notation
Before deriving the change of base formula, recall the key properties of logarithms:
- Definition: \( \log_a y = x \iff a^x = y \), where \( a \gt 0 \), \( a \ne 1 \), and \( y \gt 0 \).
- Product rule: \( \log_b(mn) = \log_b m + \log_b n \)
- Quotient rule: \( \log_b\left(\dfrac{m}{n}\right) = \log_b m - \log_b n \)
- Power rule: \( \log_b(m^p) = p \log_b m \)
- Identity: \( \log_b b = 1 \) and \( \log_b 1 = 0 \)
The power rule is the key identity used in the proof below.
Change of Base Formula Proof
Let \( y = a^x \), where \( a \gt 0 \), \( a \ne 1 \), and \( y \gt 0 \). First, convert it into its logarithmic form:
\[ \log_a y = x \qquad (1) \]
Next, take the logarithm with base \( b \) (where \( b \gt 0 \) and \( b \ne 1 \)) of both sides of the original equation \( y = a^x \):
\[ \log_b y = \log_b (a^x) \]
Use the power rule of logarithms to rewrite \( \log_b (a^x) \) as \( x \log_b a \):
\[ \log_b y = x \log_b a \]
Substitute \( x \) with \( \log_a y \) from equation (1):
\[ \log_b y = (\log_a y)(\log_b a) \]
Finally, solve for \( \log_a y \) by dividing both sides by \( \log_b a \):
Note: You can choose any base \( b \) (where \( b \gt 0 \) and \( b \ne 1 \)) to rewrite a logarithm. The most common choices are base 10 (log) and base \( e \) (ln), since these are available on most calculators.
How to Apply the Formula
- Identify the original base and argument: Write \( \log_a y \) and note \( a \) (the base) and \( y \) (the argument).
- Choose a new base \( b \): Usually base 10 or base \( e \), since those are available on most calculators.
- Apply the formula: \( \log_a y = \dfrac{\log_b y}{\log_b a} \).
- Evaluate: Compute each logarithm and divide.
Memory aid: The argument goes on top, and the original base goes on the bottom.
Examples of Applications
Example 1: Evaluating Logarithms
Show Solution
Question:
a) Evaluate \( \log_4 16 \) noting that \( 16 = 4^2 \).
b) Use the change of base formula to rewrite \( \log_4 16 \) using a logarithm with base 2, and evaluate it again. Compare the results.
Solution
a) Using the standard base:
\[ \log_4 16 = \log_4 (4^2) = 2 \]b) Using the change of base formula with base 2:
\[ \log_4 16 = \dfrac{\log_2 16}{\log_2 4} \]Note that \( 4 = 2^2 \) and \( 16 = 2^4 \). Therefore:
\[ \log_4 16 = \dfrac{\log_2 (2^4)}{\log_2 (2^2)} = \dfrac{4}{2} = 2 \]Conclusion: The evaluation in parts (a) and (b) yields the same answer, verifying the formula.
Example 2: Using the Natural Logarithm
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Question: Express \( \log_2 x \) using the natural logarithm \( \ln \).
Solution
Using the change of base formula with the natural logarithm (where base \( b = e \)), we write:
\[ \log_2 x = \dfrac{\ln x}{\ln 2} \]Note: This conversion is highly useful in calculus for calculating the derivative and integral of logarithmic functions to any base.
Example 3: Evaluating a Logarithm with an Unusual Base
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Question: Evaluate \( \log_5 100 \) using the change of base formula. Round your answer to four decimal places.
Solution
Using base 10:
\[ \log_5 100 = \dfrac{\log_{10} 100}{\log_{10} 5} = \dfrac{2}{\log_{10} 5} \]Since \( \log_{10} 5 \approx 0.69897 \):
\[ \log_5 100 \approx \dfrac{2}{0.69897} \approx 2.8614 \]Check: \( 5^{2.8614} \approx 100 \) ✓
Alternatively, using natural logarithms:
\[ \log_5 100 = \dfrac{\ln 100}{\ln 5} = \dfrac{4.60517}{1.60944} \approx 2.8614 \]Both methods produce the same result, confirming the formula works with any valid base.
Example 4: Simplifying a Ratio of Logarithms
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Question: Simplify \( \dfrac{\log_7 81}{\log_7 3} \).
Solution
Notice that this ratio matches the structure of the change of base formula in reverse. Using the formula with base \( 7 \):
\[ \log_3 81 = \dfrac{\log_7 81}{\log_7 3} \]Since \( 81 = 3^4 \):
\[ \log_3 81 = \log_3 (3^4) = 4 \]So the ratio simplifies to \( \boxed{4} \).
Insight: Any ratio of the form \( \dfrac{\log_b M}{\log_b N} \) is equal to \( \log_N M \). This is one of the most useful consequences of the change of base formula.
Example 5: Solving an Exponential Equation Using Change of Base
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Question: Solve \( 3^x = 20 \) for \( x \), giving your answer correct to four decimal places.
Solution
Take \( \log_{10} \) of both sides:
\[ \log_{10}(3^x) = \log_{10} 20 \]Apply the power rule:
\[ x \log_{10} 3 = \log_{10} 20 \]Solve for \( x \):
\[ x = \dfrac{\log_{10} 20}{\log_{10} 3} \]This is exactly \( \log_3 20 \) by the change of base formula. Evaluating:
\[ x = \dfrac{1.30103}{0.47712} \approx 2.7268 \]Check: \( 3^{2.7268} \approx 20 \) ✓
Example 6: Change of Base with a Fractional Argument
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Question: Evaluate \( \log_8 \left(\dfrac{1}{32}\right) \) exactly using the change of base formula.
Solution
Use base 2 (both \( 8 \) and \( 32 \) are powers of 2):
\[ \log_8 \left(\dfrac{1}{32}\right) = \dfrac{\log_2 (1/32)}{\log_2 8} \]Since \( \dfrac{1}{32} = 2^{-5} \) and \( 8 = 2^3 \):
\[ \log_8 \left(\dfrac{1}{32}\right) = \dfrac{-5}{3} \]Check: \( 8^{-5/3} = (2^3)^{-5/3} = 2^{-5} = \dfrac{1}{32} \) ✓
Common Mistakes to Avoid
- Inverting the fraction. Writing \( \log_a y = \dfrac{\log_b a}{\log_b y} \) is incorrect. The argument \( y \) goes on top, and the base \( a \) goes on the bottom.
- Forgetting the domain conditions. The formula requires \( a \gt 0 \), \( a \ne 1 \), \( b \gt 0 \), \( b \ne 1 \), and \( y \gt 0 \).
- Cancelling logarithms incorrectly. \( \dfrac{\log_b y}{\log_b a} \neq \log_b\!\left(\dfrac{y}{a}\right) \). The quotient rule applies to the argument of a single logarithm, not to a ratio of two logarithms.
- Choosing an invalid base. Base \( b = 1 \) is not allowed, and base \( b = 0 \) or negative bases are not valid for logarithms.
- Mixing bases on the right-hand side. Both logarithms must use the same base \( b \).
Table of Useful Logarithmic Values
The following values are handy when applying the change of base formula with base 10 or base \( e \):
| \( x \) | \( \log_{10} x \) | \( \ln x \) |
|---|---|---|
| 2 | 0.30103 | 0.69315 |
| 3 | 0.47712 | 1.09861 |
| 5 | 0.69897 | 1.60944 |
| 7 | 0.84510 | 1.94591 |
| 10 | 1.00000 | 2.30259 |
| 100 | 2.00000 | 4.60517 |
Tip: You can verify any calculation by checking that \( a^{\log_a y} = y \).
Real-World Applications
- Calculus: The derivative of \( \log_a x \) is \( \dfrac{1}{x \ln a} \), which requires converting to natural logarithms. Similarly, \( \displaystyle\int \log_a x \, dx = \dfrac{x \ln x - x}{\ln a} + C \).
- Computer science: Logarithmic complexity is often expressed as \( \log_2 n \), but when analyzing algorithms against base-\( e \) models, the change of base formula is used.
- Chemistry (pH): The pH scale uses \( \text{pH} = -\log_{10}[\text{H}^+] \). Converting to natural logarithms is common when working with thermodynamic equations.
- Finance: Compound interest and continuous growth use natural logarithms; the change of base formula lets you compare growth rates expressed in different bases.
- Information theory: Entropy can be measured in bits (base 2) or nats (base \( e \)); the change of base formula converts between them.
Frequently Asked Questions
What is the change of base formula for logarithms?
The change of base formula states that \( \log_a y = \dfrac{\log_b y}{\log_b a} \), where \( a \) is the original base, \( b \) is the new base (\( b \gt 0 \) and \( b \ne 1 \)), and \( y \) is a positive real number.
Why is the change of base formula useful?
The change of base formula is useful because most calculators only have buttons for base-10 logarithms (\( \log \)) and natural logarithms (\( \ln \)). It allows you to evaluate logarithms in any base by converting them to a base your calculator supports.
Can you use any base in the change of base formula?
Yes. You can choose any base \( b \) as long as \( b \gt 0 \) and \( b \ne 1 \). The most common choices are base 10 (\( \log \)) and base \( e \) (\( \ln \)), because these are the bases available on most calculators.
What are the conditions for the change of base formula?
The change of base formula requires: (1) the argument \( y \gt 0 \), (2) the original base \( a \gt 0 \) and \( a \ne 1 \), and (3) the new base \( b \gt 0 \) and \( b \ne 1 \).
How is the change of base formula derived?
Start with \( y = a^x \), take the logarithm base \( b \) of both sides to get \( \log_b y = x \log_b a \). Since \( x = \log_a y \), substitute to obtain \( \log_b y = \log_a y \cdot \log_b a \). Dividing both sides by \( \log_b a \) gives \( \log_a y = \dfrac{\log_b y}{\log_b a} \).
What is a common mistake when using the change of base formula?
A common mistake is writing \( \log_a y = \dfrac{\log_b a}{\log_b y} \), which inverts the fraction. The correct formula places the log of the argument \( y \) in the numerator and the log of the original base \( a \) in the denominator.
Can the change of base formula be used to simplify ratios of logarithms?
Yes. Any ratio of the form \( \dfrac{\log_b M}{\log_b N} \) equals \( \log_N M \). This is a direct consequence of the change of base formula and is often used to simplify complex logarithmic expressions.
More References and Links
- Convert Logarithms and Exponentials
- Rules of Logarithm and Exponential - Questions with Solutions
- Derivative of Logarithmic Functions
- Integral of Logarithmic Functions
- Logarithmic Functions
- Precalculus - Mathematics for Calculus - 7th Edition - James Stewart, Lothar Redlin, Saleem Watson
- Algebra and Trigonometry - 10th Edition - Michael Sullivan