Finding the Logarithm Value of a Product

We are asked to find the value of a product involving several logarithm terms: $(\log_3 4) (\log_4 5) (\log_5 6) (\log_6 7) (\log_7 8) (\log_8 9)$. To solve this, we can use a fundamental property of logarithms called the logarithm base change formula. This formula is very useful for simplifying products like this one and determining the final logarithm value.

Using the Logarithm Base Change Formula

The base change formula states that $\log_b a = \frac{\log_c a}{\log_c b}$, where $c$ can be any convenient base. Using this formula, we can rewrite each term in the product:

• $\log_3 4 = \frac{\log 4}{\log 3}$
• $\log_4 5 = \frac{\log 5}{\log 4}$
• $\log_5 6 = \frac{\log 6}{\log 5}$
• $\log_6 7 = \frac{\log 7}{\log 6}$
• $\log_7 8 = \frac{\log 8}{\log 7}$
• $\log_8 9 = \frac{\log 9}{\log 8}$

Here, $\log$ without a subscript represents the logarithm with respect to any common base, as it will cancel out when multiplied.

Evaluating the Logarithm Product

Now, let's substitute these expressions back into the original logarithm product:

$(\log_3 4) (\log_4 5) (\log_5 6) (\log_6 7) (\log_7 8) (\log_8 9) = \left(\frac{\log 4}{\log 3}\right) \times \left(\frac{\log 5}{\log 4}\right) \times \left(\frac{\log 6}{\log 5}\right) \times \left(\frac{\log 7}{\log 6}\right) \times \left(\frac{\log 8}{\log 7}\right) \times \left(\frac{\log 9}{\log 8}\right)$

Notice the pattern of cancellation. The numerator of each fraction cancels with the denominator of the subsequent fraction:

$= \frac{\cancel{\log 4}}{\log 3} \times \frac{\cancel{\log 5}}{\cancel{\log 4}} \times \frac{\cancel{\log 6}}{\cancel{\log 5}} \times \frac{\cancel{\log 7}}{\cancel{\log 6}} \times \frac{\cancel{\log 8}}{\cancel{\log 7}} \times \frac{\log 9}{\cancel{\log 8}}$

After all the cancellations, we are left with:

$= \frac{\log 9}{\log 3}$

Using the logarithm base change formula in reverse, $\frac{\log_c a}{\log_c b} = \log_b a$, we can simplify this expression:

$= \log_3 9$

Calculating the Final Logarithm Value

To find the logarithm value of $\log_3 9$, we need to determine the power to which the base 3 must be raised to get 9. Let this value be $x$.

$3^x = 9$

Since $9 = 3^2$, we have:

$3^x = 3^2$

Therefore, $x = 2$.

The final logarithm value obtained for the given logarithm product is 2.

This step-by-step process using logarithm properties allows us to efficiently calculate the logarithm value in such problems.