Change of Base Formula
Introduction
The change of base formula in logarithms helps us switch the base of a logarithm. For example: In calculators, we have buttons for "`log`" and "`ln`”, which help us to switch between base `10` and base e logarithms. But what if we need a logarithm with a different base? That's where the change of base formula comes in handy. It helps us in changing the base of a logarithmic term into the base we need. Plus, it's super helpful for solving complex logarithm problems involving terms having different base values. So, let's try to understand the formula, and how it works, and work on a few examples to get the hang of it.
Understanding Change of Base Formula
The change of base formula is a math trick used with logarithms. It helps when you want to change the base of a logarithm to a different one. Let's say you have a logarithm with a base, but you want to find it with a different base. That's where we apply the change of base formula. It's like switching gears to work with different bases. The change of base formula is:
`\log_{b} a = \frac{\log_{c} a}{\log_{c} b}`
In this formula:
- The number inside the logarithm in the top part (numerator) is the same as the original number you were working with.
- The number inside the logarithm in the bottom part (denominator) is the same as the original base of the logarithm.
- Both logarithms, in the top and bottom parts, must have the same base, which can be any positive number except `1`.
Another way to express this formula is: `\log_{b} a \cdot \log_{c} b = \log_{c} a`
Derivation of Change of Base Formula
We start with the assumption that we have a logarithm base ` b ` of ` a `, denoted as ` \log_{b} a `, and another logarithm base ` c ` of the same number ` a `, denoted as ` \log_{c} a `, and we want to find a relationship between them.
Let us assume that:
`\log_{b} a = p, \quad \log_{c} a = q, \quad \text{and} \quad \log_{c} b = r`
Now, let's convert these logarithmic equations into exponential form:
`a = b^p, \quad a = c^q, \quad \text{and} \quad b = c^r`
From the first two equations, we get:
`b^p = c^q`
Substituting ` b = c^r ` from the third equation, we have:
`(c^r)^p = c^q`
Using the property of exponents `(a^m)^n = a^{mn}`, we simplify to:
`c^{rp} = c^q`
Which leads to:
\( rp = q \)
Finally, rearranging this equation, we get:
`p = \frac{q}{r}`
Substituting the values of ` p `, ` q `, and ` r `, we have:
`\log_{b} a = \frac{\log_{c} a}{\log_{c} b}`
Key Points Regarding Change of `Log` Base
- The change of base rule in logarithms is a valuable property allowing the use of logarithms with bases other than `10` and `e` in calculators.
- This rule is represented by the equation: ` \log_{d} c = \frac{\log_{10} c}{\log_{10} d} `.
- Utilizing the base change formula is particularly handy when working with calculators that solely support base `10` and base `e` logarithms.
- Notably, the base change formula extends its utility to the natural logarithm, ` \ln `, through the equation: ` \log b = \frac{\ln b}{\ln a} `.
- It's essential to note that the formula applies exclusively to logarithms with positive bases.
- Both the numerator and the denominator of the formula signify logarithms with the identical base ` c `, facilitating seamless calculations and interpretation.
- When the base of a logarithm is not mentioned, as in ` \log `, we implicitly consider the base to be `10`. This means that ` \log x ` and ` \log_{10} x ` are the same thing.
Solved Examples
Example `1`: Determine ` \log_{49} 7 ` using the change of base formula.
Solution:
Using the change of base formula, we rewrite the given logarithm in terms of base `10`.
(Note that ` \log ` is the same as ` \log_{10}`).
`\log_{49} 7 = \frac{\log 7}{\log 49}`
`= \frac{\log 7}{\log (7^2)}`
`= \frac{\log 7}{2 \log 7} \quad \text{(since } \log (a^m) = m \log a \text{)}`
`= \frac{1}{2}`
So, `\log_{49} 7 = \frac{1}{2}`.
Example `2`: Evaluate ` \log_{3} 25 ` using the change of base formula. Round your answer to `3` decimals.
Solution:
Using the change of base formula, we rewrite the given logarithm in terms of base `10`:
`\log_{3} 25 = \frac{\log_{10} 25}{\log_{10} 3}`
Now, we calculate the values of ` \log_{10} 25 ` and ` \log_{10} 3 ` using a calculator:
\( \log_{10} 25 \approx 1.3979 \)
\( \log_{10} 3 \approx 0.4771 \)
Substituting these values into the formula, we have:
`\log_{3} 25 \approx \frac{1.3979}{0.4771}`
`\approx 2.929`
So, \( \log_{3} 25 \approx 2.929 \).
Example `3`: Determine ` \log_{4} 64 ` using the change of base formula.
Solution:
Using the change of base formula, we rewrite the given logarithm in terms of base `10`.
(Note that ` \log ` is the same as ` \log_{10}`).
`\log_{4} 64 = \frac{\log 64}{\log 4}`
`= \frac{\log (4^3)}{\log 4}`
`= \frac{3 \log 4}{\log 4} \quad \text{(since } \log (a^m) = m \log a \text{)}`
`= 3`
So, \( \log_{4} 64 = 3 \)
Example `4`: Find ` \log_{5} 100 ` using the change of base formula. Round your answer to `4` decimals.
Solution:
Using the change of base formula, we rewrite the given logarithm in terms of base `10`:
`\log_{5} 100 = \frac{\log_{10} 100}{\log_{10} 5}`
Now, we calculate the values of ` \log_{10} 100 ` and ` \log_{10} 5 ` using a calculator:
\( \log_{10} 100 = 2 \)
\( \log_{10} 5 \approx 0.6989 \)
Substituting these values into the formula, we have:
`\log_{5} 100 \approx \frac{2}{0.6989}`
`\approx 2.8616`
So, \( \log_{5} 100 \approx 2.8616 \).
Example `5`: Evaluate ` \log_{2} 5 \cdot \log_{5} 3 \cdot \log_{6} 2`.
Solution:
\( \log_{2} 5 \cdot \log_{5} 3 \cdot \log_{6} 2 \)
Considering the first two terms ` \log_{2} 5 \cdot \log_{5} 3`, we can simplify:
\( \log_{2} 5 \cdot \log_{5} 3 \)
`= \frac{log_{10} 5}{log_{10} 2} \cdot \frac{log_{10} 3}{log_{10} 5}`
`= \frac{log_{10} 3}{log_{10} 2}`
\( = \log_{2} 3 \)
Therefor `\log_{2} 5 \cdot \log_{5} 3 \cdot \log_{6} 2` now becomes `\log_{2} 3 \cdot \log_{6} 2`.
We can rewrite the expression using the multiplication property of logarithms:
`\log_{b} a \cdot \log_{c} b = \log_{c} a`
`\log_{2} 3 \cdot \log_{6} 2 = \log_{6} 3`
Therefore, ` \log_{2} 5 \cdot \log_{5} 3 \cdot \log_{6} 2 = \log_{6} 3 `.
Practice Problems
Q`1`. Determine ` \log_{81} 9 ` using the change of base formula.
- \( 2 \)
- `\frac{1}{2}`
- `\frac{1}{4}`
- \( 4 \)
Answer: b
Q`2`. Evaluate ` \log_{2} 8 ` using the change of base formula.
- \( 2 \)
- \( 3 \)
- \( 4 \)
- \( 5 \)
Answer: b
Q`3`. Evaluate ` \log_{11} 10 ` using the change of base formula. Round your answer to `4` decimals.
- \( 0.9602 \)
- \( 0.9313 \)
- \( 0.8575 \)
- \( 0.8134 \)
Answer: a
Q`4`. Find ` \log_{4} 6 \cdot \log_{6} 7 ` using the change of base formula.
- \( \log_{7} 4 \)
- \( \log_{4} 28 \)
- \( \log_{7} 28 \)
- \( \log_{4} 7 \)
Answer: d
Q`5`. Find ` \log_{8} 9 ` using the change of base formula. Round your answer to `4` decimals.
- \( 0.7924 \)
- \( 0.8982 \)
- \( 1.0566 \)
- \( 0.9375 \)
Answer: c
Frequently Asked Questions
Q`1`. What is the change of base formula in logarithms?
Answer: The change of base formula in logarithms allows us to convert a logarithm with a certain base ` b ` to a logarithm with a different base ` c `. It is represented as:
`\log_{c} a = \frac{\log_{b} a}{\log_{b} c}`
Q`2`. Why is the change of base formula useful?
Answer: The change of base formula is valuable because it enables us to work with logarithms of bases other than `10` and `e`, which is commonly supported in calculators. This versatility expands the scope of problems we can solve using logarithmic functions.
Q`3`. Can the change of base formula be used with any logarithmic expression?
Answer: Yes, the change of base formula can be applied to any logarithmic expression where the base is positive and not equal to `1`. However, it's important to ensure that the expression is defined for the given inputs.
Q`4`. Are there any restrictions on using the change of base formula?
Answer: Yes, the change of base formula can only be applied to logarithms with positive bases. Additionally, the expression within the logarithm must be positive.
Q`5`. How does the change of base formula simplify complex logarithmic expressions?
Answer: The change of base formula simplifies complex logarithmic expressions by allowing us to rewrite them in terms of logarithms with bases that are easier to work with or are supported by calculators. This simplification often leads to more straightforward computations and clearer interpretations of results.