Question

Find the value of \(\log _{2} 3 \cdot \log _{3} 4 \cdot \cdots \cdot \log _{n}(n+1) \cdot \log _{n+1} 2\)

Short answer

The value of the expression is 1.

Step-by-step solution

Understand the Given Expression

The given expression is a product of logarithms: \[\text{Expression} = \log _{2} 3 \cdot \log _{3} 4 \cdot \cdots \cdot \log _{n}(n+1) \cdot \log _{n+1} 2\]. The task is to find its value.

Use the Change of Base Formula

Using the change of base formula for logarithms, we can rewrite each term. The change of base formula is \[\text{log}_a b = \frac{\text{log}_c b}{\text{log}_c a}\]. Let's apply this to each term.

Simplify Each Logarithm

Notice that: \[\log_{2} 3 = \frac{\log 3}{\log 2}, \log _{3} 4 = \frac{\log 4}{\log 3}, \ldots, \log _{n}(n+1) = \frac{\log (n+1)}{\log n}, \log_{n+1} 2 = \frac{\log 2}{\log (n+1)}. \]Thus, the product becomes: \[\frac{\log 3}{\log 2} \cdot \frac{\log 4}{\log 3} \cdot \cdots \cdot \frac{\log (n+1)}{\log n} \cdot \frac{\log 2}{\log (n+1)}.\]

Cancel Out the Logarithms

Notice that in the product, all the logarithms in the numerator and denominator will cancel out each other except the first and the last. This results in: \[\frac{\log 3}{\log 2} \cdot \frac{\log 4}{\log 3} \cdot \cdots \cdot \frac{\log (n+1)}{\log n} \cdot \frac{\log 2}{\log (n+1)} = \frac{\log 2}{\log 2} = 1.\]

Key concepts

Change of Base Formula

When working with logarithms, you often need to change the base to make calculations easier or fit a specific need.
The change of base formula is a crucial tool for this: \[\text{log}_a b = \frac{\text{log}_c b}{\text{log}_c a}\]
This formula allows you to convert a logarithm from one base to another base of your choice, usually one that is easier to work with, like base 10 or base e (natural logarithm).

For example, if you have \[\text{log}_2 3\], you can rewrite it using base 10 or base e by following the change of base formula:
\[\text{log}_2 3 = \frac{\text{log}_{10} 3}{\text{log}_{10} 2}\]
This becomes highly useful in both theoretical analysis and practical computations.
In our original problem, each term uses the change of base formula to convert the logarithms.
So, \[\text{log}_2 3\] becomes \[\frac{\text{log}_{10} 3}{\text{log}_{10} 2}\], \[\text{log}_3 4\] becomes \[\frac{\text{log}_{10} 4}{\text{log}_{10} 3}\], and so on.
By consistently applying this formula, we pave the way for further simplification.

Logarithm Simplification

Logarithms can often be simplified by leveraging their properties.
After applying the change of base formula, our expression looked like this:
\[\frac{\text{log}_{10} 3}{\text{log}_{10} 2} \times \frac{\text{log}_{10} 4}{\text{log}_{10} 3} \times \frac{\text{log}_{10} 5}{\text{log}_{10} 4} \times \text{\textellipsis} \times \frac{\text{log}_{10} 2}{\text{log}_{10} (n+1)}\]
Notice that each \[\text{log}_{10} b\] term in the numerator of one fraction cancels with the denominator of the next.

This specific setup allows for cross-cancellation, ultimately leaving:
\[ \frac{\text{log}_{10} 2}{\text{log}_{10} 2} = 1 \]
Whether you are working with numbers or more complex mathematical expressions, this approach simplifies the computation and makes it possible to evaluate expressions.

Logarithmic Product

Understanding the product of logarithms involves knowing how logarithmic expressions interact.
When you multiply logarithms, each term can often be simplified or canceled based on properties of the logarithms.

In our specific exercise, the product starts as:
\[\text{log}_2 3 \times \text{log}_3 4 \times \text{log}_4 5 \times \text{\textellipsis} \times \text{log}_{n}(n+1) \times \text{log}_{n+1} 2\]
This transforms, via the change of base formula, into a sequence of fractions:
\[ \frac{\text{log}_{10} 3}{\text{log}_{10} 2} \times \frac{\text{log}_{10} 4}{\text{log}_{10} 3} \times \text{\textellipsis} \times \frac{\text{log}_{10} 2}{\text{log}_{10} (n+1)} \]
Finally, these fractions simplify to leave \[\frac{\text{log}_{10} 2}{\text{log}_{10} 2} = 1\]
Thus, the product of these logarithms is 1.

This exercise highlights how understanding the interactions within logarithmic products is essential.
The ability to simplify complex logarithmic expressions down to a basic form exemplifies the power and importance of mastering logarithm rules.