Question

Find the value of \(\log _{2} 3 \cdot \log _{3} 4 \cdot \log _{4} 5 \cdot \log _{5} 6 \cdot \log _{6} 7 \cdot \log _{7} 8\).

Short answer

3

Step-by-step solution

- Understand the Logarithm Chain

Recognize that the expression \(\log_{2} 3 \cdot \log_{3} 4 \cdot \log_{4} 5 \cdot \log_{5} 6 \cdot \log_{6} 7 \cdot \log_{7} 8\) is a product of logarithms with changing bases and internal arguments which create a chain pattern.

- Apply the Change of Base Formula

Use the change of base formula for logarithms: \(\log_{a} b = \frac{\log b}{\log a}\). Apply this to each term in the product: \( \log_{2}3 = \frac{\log3}{\log2}, \log_{3}4 = \frac{\log4}{\log3}, \dots, \log_{7}8 = \frac{\log8}{\log7}\).

- Substitute and Simplify

Substitute the expressions obtained from the change of base formula into the original product: \[ \frac{\log 3}{\log 2} \cdot \frac{\log 4}{\log 3} \cdot \frac{\log 5}{\log 4} \cdot \frac{\log 6}{\log 5} \cdot \frac{\log 7}{\log 6} \cdot \frac{\log 8}{\log 7} \]

- Cancel Out Logarithms

Notice that each \(\log\) term in the numerators and denominators will cancel with the adjacent term: \[ \frac{\log 3}{\log 2} \cdot \frac{\log 4}{\log 3} \cdot \frac{\log 5}{\log 4} \cdot \frac{\log 6}{\log 5} \cdot \frac{\log 7}{\log 6} \cdot \frac{\log 8}{\log 7} = \frac{\log 8}{\log 2} \]

- Simplify the Final Expression

The remaining logarithm expression is \(\frac{\log 8}{\log 2}\). Apply the logarithm properties: \[ \log_{2} 8 = \frac{\log 8}{\log 2} \]

- Calculate the Final Value

Recall that 8 is \(2^3\), so \(\log_{2} 8 = \log_{2} (2^3) = 3\).

Key concepts

logarithm properties

A logarithm is an exponent that indicates the power to which a base number must be raised to get a certain value. Understanding properties of logarithms can make solving problems easier. Here are some key properties:

• Product Property: \ \( \log_b(xy) = \log_b(x) + \log_b(y) \)
• Quotient Property: \ \( \log_b(\frac{x}{y}) = \log_b(x) - \log_b(y) \)
• Power Property: \ \( \log_b(x^k) = k \log_b(x) \)

These properties help simplify logarithmic expressions. For example, knowing that \ \( \log_{10}(1000) \) is just an exponent, with the base of 10 raised to the power of 3 to get 1000, we can directly determine that \ \( \log_{10}(1000) = 3 \). This directly aids in simplifying complex expressions, such as our given problem.

change of base formula

The change of base formula is another useful tool when dealing with logarithms. It allows you to convert a logarithm of any base to a different base. The formula is: \ \( \log_{a}(b) = \frac{\log_{c}(b)}{\log_{c}(a)} \). This is particularly helpful if you have a calculator that only handles base 10 or base e (natural logarithms).
For instance, in our example, to solve expressions like \ \( \log_{2}(3) \), you could convert it using a natural logarithm as follows: \ \( \log_{2}(3) = \frac{\ln(3)}{\ln(2)} \) or base 10 logarithm: \ \( \log_{2}(3) = \frac{\log(3)}{\log(2)} \). This ability to switch bases makes it easier to calculate logarithms and is key to simplifying the chained logarithmic expressions piece by piece.

algebraic simplification

Algebraic simplification aids in reducing complex logarithmic expressions into simpler forms that are easier to evaluate. By applying the change of base formula, the initial problem transforms like this:
\ \( \frac{\log(3)}{\log(2)} \cdot \frac{\log(4)}{\log(3)} \cdot \frac{\log(5)}{\log(4)} \cdot \frac{\log(6)}{\log(5)} \cdot \frac{\log(7)}{\log(6)} \cdot \frac{\log(8)}{\log(7)} \)
Then, simplify by canceling out terms in the numerator and denominator, pairing each logarithm with its adjacent opposite:
\ \( \frac{\log(3)}{\log(2)} \cdot \frac{\log(4)}{\log(3)} \rightarrow \cdots \rightarrow \ \frac{\log(8)}{\log(2)} \).
This reduces to \ \( \log_{2}(8) \), which simplifies further knowing \ \( 8 = 2^3 \), leading to the final result: 3. This approach underscores how breaking expressions down step-by-step makes solving them manageable.