Problem 94 - Intermediate Algebra

Question

Assume that \(\log 4 \approx 0.6021, \log 7 \approx 0.8451,\) and \(\log 9 \approx 0.9542 .\) Use these values to evaluate each logarithm. See Example 8 . $$ \log _{b} 49 $$

Step-by-Step Solution

Answer

\(\log 49 \approx 1.6902\)

Step 1: Convert to Logarithmic Form Using Change of Base Formula

To evaluate \( \log_b 49 \) using known logarithms of base 10, use the change of base formula: \( \log_b 49 = \frac{\log 49}{\log b} \). However, we aren't given \(\log b\), so we'll assume \(b = 10\), which makes our task to evaluate \(\log_{10} 49\).

Step 2: Express 49 as a Product of Known Values

Notice that 49 can be expressed as \( 7^2 \). Hence, \( \log 49 = \log (7^2) \).

Step 3: Apply the Power Rule of Logarithms

By using the power rule, \( \log (7^2) = 2 \log 7 \). This allows us to leverage the known value of \( \log 7 \).

Step 4: Substitute Known Logarithmic Values

Substitute \( \log 7 \approx 0.8451 \) into the equation: \( 2 \log 7 \approx 2 \times 0.8451 = 1.6902 \).

Step 5: Conclude the Result

Thus, \( \log 49 \) is approximately equal to \(1.6902\). This is the value of \(\log_{10} 49\).

Key Concepts

Change of Base Formula

When dealing with logarithms, the change of base formula is an essential tool that allows you to switch the bases of logarithms. This is particularly useful when you don't know the base or when a specific base calculator is not available. The formula is given by: \[ \log_b a = \frac{\log_c a}{\log_c b} \] where \( b \) is the original base, \( a \) is the number we are taking the logarithm of, and \( c \) is the new base.

• You can choose any base \( c \) you like, but common choices are 10 or \( e \) for natural logarithms.
• In many cases, calculators don't have a log button for every base, only for base 10 (common logs) and \( e \) (natural logs), making this formula even more valuable.

For example, if you want to calculate \( \log_2 8 \), you might switch to base 10: \[ \log_2 8 = \frac{\log_{10} 8}{\log_{10} 2} \] which you can then easily evaluate using a calculator.

Power Rule of Logarithms

The power rule of logarithms is a powerful property that simplifies logarithms involving exponents. It states that \( \log_b (a^n) = n \cdot \log_b a \).

• This rule is derived from the fundamental property of exponents, as it transforms multiplication into a manageable addition task.
• It allows you to "bring down" the exponent to make calculations simpler.

For instance, if you want to calculate \( \log (7^2) \), you can apply the power rule, leading to \( 2 \times \log 7 \). In our exercise, this step was critical because we could then use the known value of \( \log 7 \approx 0.8451 \).Thus, \( \log 49 = 2 \times 0.8451 = 1.6902 \), making it much easier to compute than directly handling \( 49 \).

Base 10 Logarithms

Base 10 logarithms, also known as common logarithms, have several practical uses. They are often written simply as \( \log \) rather than \( \log_{10} \).

• This base is particularly prevalent in science and engineering, where it helps to deal with measurements of scale, like decibels, pH levels, or Richter magnitudes.
• Since calculators typically have a button for base 10 logarithms, they are convenient for computational problems.

In our provided solution, we simplified the problem by assuming a base 10 logarithm to facilitate the calculation. This is because a base 10 allows us to directly use known values or a calculator, resulting in a quick and efficient solution.

Evaluating Logarithms

Evaluating logarithms can be intimidating, but understanding the right steps makes it much easier. The evaluation typically involves:

• Identifying if the logarithm can be simplified using properties such as the power, product, or quotient rules.
• Converting complex bases to base 10 using the change of base formula whenever possible, as discussed previously.
• Using known logarithmic values and operations to compute the result.For our exercise, evaluating \( \log_{10} 49 \) was simpler due to expressing it as \( \log (7^2) \) and applying the power rule.

By substituting known values, such as \( \log 7 \approx 0.8451 \), we used arithmetic to determine that \( \log 49 \approx 1.6902 \). This showcases how cleverly applying rules and expressions helps to break down what initially seems to be a complex problem.