Problem 34

Question

Use \(\log _{a} x=(\ln x) /(\ln a)\) to calculate each of the logarithms in Problems 33-36. \(\log _{7}(0.11)\)

Step-by-Step Solution

Verified
Answer
\(\log_7(0.11) \approx -1.13402\).
1Step 1: Identify the Conversion Formula
We need to use the formula to convert the logarithm to a natural logarithm format. The formula given is \(\log _{a} x= \frac{\ln x}{\ln a}\). This tells us we need to convert the expression \(\log_{7}(0.11)\) using natural logarithms.
2Step 2: Apply the Conversion Formula
Substitute \(x = 0.11\) and \(a = 7\) into the formula. This gives us: \[ \log_7(0.11) = \frac{\ln(0.11)}{\ln(7)} \].
3Step 3: Compute the Natural Logarithm of 0.11
Using a calculator, find \(\ln(0.11)\). This value is approximately \(-2.20727\).
4Step 4: Compute the Natural Logarithm of 7
Next, find \(\ln(7)\) using a calculator. This value is approximately \(1.94591\).
5Step 5: Divide the Natural Logarithms to Find the Result
Divide the result from Step 3 by the result from Step 4: \[ \frac{-2.20727}{1.94591} \approx -1.13402 \]. This is the value of \(\log_7(0.11)\).

Key Concepts

Understanding Natural LogarithmsThe Conversion Formula ExplainedMastering Base Change for Logarithms
Understanding Natural Logarithms
Natural logarithms are logarithms with the base of the mathematical constant \( e \), which is approximately 2.71828. The symbol for a natural logarithm is \( \ln \), and it is used to solve equations involving exponential growth and decay, among many other applications.
Natural logarithm is often used because the base \( e \) appears frequently in mathematical modeling of natural phenomena. It simplifies many complex calculations and makes it easier to solve various equations. Keep in mind that \( \ln(x) \) is only defined for positive numbers because you cannot take the logarithm of a negative number or zero.
  • Natural logarithms are logarithms with base \( e \)
  • Symbolized by \( \ln \)
  • Commonly used in calculus and exponential functions
Understanding how to work with \( \ln \) is key to converting between different types of logarithmic bases.
The Conversion Formula Explained
The conversion formula is essential when you want to change the base of a logarithm to another base. The formula, \( \log_{a}(x) = \frac{\ln(x)}{\ln(a)} \), allows you to convert any logarithm with base \( a \) into a natural logarithm. This is extremely helpful when you need to compare or compute logarithms that involve different bases.
In our exercise, we use this conversion formula to express \( \log_7(0.11) \) in terms of natural logarithms. We replace \( a = 7 \) and \( x = 0.11 \) in the formula to get \( \log_7(0.11) = \frac{\ln(0.11)}{\ln(7)} \). By using natural logarithms, calculations become more straightforward because it's simpler to use \( \ln \) with calculators and various mathematical software.
  • The conversion formula is \( \log_{a}(x) = \frac{\ln(x)}{\ln(a)} \)
  • Allows conversion to base \( e \)
  • Simplifies computing and comparing different logarithmic bases
Understanding this conversion lays the foundation for the following concept, which is changing the base of logarithms.
Mastering Base Change for Logarithms
The concept of changing the base of logarithms hinges on using the conversion formula we discussed earlier. This process is useful because it lets us compute logarithms that may not be easily accessible with a particular base.
The conversion formula plays a crucial role here. By transforming \( \log_a(x) \) into a natural logarithm format, we can more easily handle these calculations using calculators, which primarily support natural logarithms (\( \ln \)) and base 10 logarithms (\( \log_{10} \)).
Let's revisit the steps from the exercise:
  • Identify the need for base change using the conversion formula.
  • Apply the formula by substituting the actual values you have.
  • Use a calculator to find the natural logarithm values needed.
  • Perform the division to get your result.
By mastering these steps, you will be equipped to handle any logarithmic problem that involves base changes proficiently.