If (log3 4)(log4 5)(log5 6) · · · (log30 31)(log31 32) = a/b log2(3) , where a and b a positive integers
with no common factors, determine the value of a + b.

We are given the product:

\[
(\log_3 4)(\log_4 5)(\log_5 6) \cdots (\log_{30} 31)(\log_{31} 32)
\]

and we need to simplify this expression and express it in the form \(\frac{a}{b} \log_2(3)\), where \(a\) and \(b\) are coprime integers, and then find the value of \(a + b\).

### Step 1: Using the Change of Base Formula

We can use the change of base formula for logarithms, which states:

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

We rewrite each term \(\log_{n}(n+1)\) using the natural logarithm (\(\log\) can refer to any logarithmic base, but we'll assume it's base 10):

\[
\log_n (n+1) = \frac{\log(n+1)}{\log(n)}
\]

Thus, the product becomes:

\[
(\log_3 4)(\log_4 5)(\log_5 6) \cdots (\log_{31} 32) = \frac{\log 4}{\log 3} \times \frac{\log 5}{\log 4} \times \frac{\log 6}{\log 5} \times \cdots \times \frac{\log 32}{\log 31}
\]

### Step 2: Simplifying the Product

Notice that in this product, many terms cancel out. Specifically, every \(\log n\) in the denominator cancels with the \(\log n\) in the numerator of the next term. The only terms that don't cancel are \(\log 32\) in the numerator and \(\log 3\) in the denominator.

Thus, the entire product simplifies to:

\[
\frac{\log 32}{\log 3}
\]

### Step 3: Expressing the Logarithms in Base 2

Now, express the logarithms in terms of base 2. We know that:

\[
\log 32 = \log_2 32 \cdot \log 2 = 5 \log 2 \quad \text{(since } 32 = 2^5\text{)}
\]

Thus:

\[
\frac{\log 32}{\log 3} = \frac{5 \log 2}{\log 3}
\]

### Step 4: Comparing with \(\frac{a}{b} \log_2(3)\)

We are given that this expression equals \(\frac{a}{b} \log_2 3\), and we know that \(\log_2 3 = \frac{\log 3}{\log 2}\). So:

\[
\frac{5 \log 2}{\log 3} = \frac{a}{b} \cdot \frac{\log 2}{\log 3}
\]

Canceling \(\frac{\log 2}{\log 3}\) on both sides, we get:

\[
5 = \frac{a}{b}
\]

Thus, \(a = 5\) and \(b = 1\).

### Step 5: Finding \(a + b\)

Finally, \(a + b = 5 + 1 = 6\).

Therefore, the value of \(a + b\) is:

\[
\boxed{6}
\]

Would you like further details or have any questions?

Here are 5 related questions to explore:

1. How does the change of base formula for logarithms work?
2. What are the properties of logarithms that simplify products of log terms?
3. How can logarithmic identities be applied in complex expressions?
4. Can you explain how natural logarithms differ from logarithms in other bases?
5. How do you interpret logarithmic equations in terms of exponents?

**Tip:** Whenever dealing with logarithmic products, consider canceling terms by rewriting the logs in a common base!

If (log3 4)(log4 5)(log5 6) · · · (log30 31)(log31 32) = a/b log2(3) , where a and b are positive integers with no common factors, determine the value of a + b.

Solve the complex logarithmic expression (log3 4)(log4 5)...(log31 32) and express it in the form a/b log2(3). Learn about the change of base formula and logarithmic properties. Suitable for Grades 10-12 students.

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