Problem 290 - Precalculus

Question

Does \(\log _{81}(2401)=\log _{3}(7) ?\) Verify the claim algebraically.

Step-by-Step Solution

Answer

Yes, the equality holds: \(\log_{81}(2401)=\log_{3}(7)\).

Step 1: Recall the Change of Base Formula

The Change of Base formula is useful in converting logarithms from one base to another. The formula is: \[\log_b(a) = \frac{\log_c(a)}{\log_c(b)}\]We can use this formula to convert both logarithms to the same base, allowing us to compare them.

Step 2: Convert \(\log_{81}(2401)\) to Base 3

To compare \(\log_{81}(2401)\) and \(\log_{3}(7)\), convert \(\log_{81}(2401)\) to base 3 using the Change of Base formula:\[\log_{81}(2401) = \frac{\log_{3}(2401)}{\log_{3}(81)}\]Recognize that \(81 = 3^4\), so: \(\log_{3}(81) = 4\).Thus,\[\log_{81}(2401) = \frac{\log_{3}(2401)}{4}\]

Step 3: Convert \(2401\) to Base 3 Terms

Note that 2401 is a power of 7:\[2401 = 7^4\]Thus,\[\log_{3}(2401) = \log_{3}(7^4) = 4 \cdot \log_{3}(7)\]

Step 4: Simplify \(\frac{\log_{3}(2401)}{4}\)

Substitute \(\log_3(2401) = 4 \cdot \log_3(7)\) into the expression:\[\frac{\log_{3}(2401)}{4} = \frac{4 \cdot \log_{3}(7)}{4} = \log_{3}(7)\]This shows that \(\log_{81}(2401) = \log_{3}(7)\).

Step 5: Conclusion: Verify the Equality

Our algebraic manipulation shows that converting \(\log_{81}(2401)\) to base 3 results in \(\log_{3}(7)\). Therefore, the claim that \(\log_{81}(2401) = \log_{3}(7)\) is true.

Key Concepts

Change of Base Formula

When working with logarithms, sometimes you may need to compare logarithms with different bases. The Change of Base formula is a handy tool for converting logarithms from one base to another, making comparison possible. The Change of Base formula is written as follows:

\[\log_b(a) = \frac{\log_c(a)}{\log_c(b)}\]

In this formula, \(b\) is the original base of the logarithm you want to convert, \(a\) is the argument, and \(c\) is the new base you will be converting to, which is often 10 or \(e\) for computational purposes. This formula allows you to express \(\log_b(a)\) in terms of a different base, simplifying comparisons between logs with different bases.

Algebraic Manipulation

Algebraic manipulation is a critical skill when dealing with logarithms. It involves rewriting expressions to make them easier to solve or compare. In the original exercise, we used algebraic manipulation to convert \(\log_{81}(2401)\) into a comparable form with \(\log_3(7)\).

Here's how it works. First, we apply the Change of Base formula:

• \(\log_{81}(2401) = \frac{\log_3(2401)}{\log_3(81)}\)
• Recognize that 81 is \(3^4\), therefore, \(\log_3(81) = 4\)
• Substitute and simplify: \(\log_{81}(2401) = \frac{\log_3(2401)}{4}\)

Further, expressing 2401 as \(7^4\), allows you to use the power rule of logs:

• \(\log_3(2401) = \log_3(7^4) = 4 \cdot \log_3(7)\)

Plug back into the earlier expression getting:

• \(\frac{4 \cdot \log_3(7)}{4} = \log_3(7)\)

Through these steps, algebraic manipulation simplifies the original logarithmic expression to confirm the equality.

Logarithmic Identities

Logarithmic identities are a set of rules that make solving logarithmic equations much simpler. These identities can help in simplifying or solving logarithmic expressions. In the original exercise, the application of logarithmic identities was pivotal to checking the given claim.

Some common logarithmic identities include:

• The Power Rule: \(\log_b(a^n) = n \cdot \log_b(a)\)
• The Product Rule: \(\log_b(xy) = \log_b(x) + \log_b(y)\)
• The Quotient Rule: \(\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)\)

In the problem, we specifically used the Power Rule when converting \(\log_3(2401)\). Recognizing that 2401 equals \(7^4\), the Power Rule allowed us to say:

• \(\log_3(7^4) = 4 \cdot \log_3(7)\)

Using these identities not only simplifies the problem but also solidifies our understanding of the properties of logarithms.