6) log_3 2 · log_4 3 · log_5 4 · ... · log_{15} 14 · log_{16} 15 =

This expression involves a sequence of logarithmic terms, each with a form [tex]\log_b a[/tex], spanning from [tex]\log_3 2[/tex] to [tex]\log_{16} 15[/tex]. These terms can be tackled using the change of base formula for logarithms, which states:

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

For simplicity and without loss of generality, we'll assume a common logarithm (either base 10 or base [tex]e[/tex], but it cancels out), so:

[tex]\log_3 2 = \frac{\log 2}{\log 3}[/tex]
[tex]\log_4 3 = \frac{\log 3}{\log 4}[/tex]

Notice that in this sequence:

Each subsequent term [tex]\log_{n+1} n[/tex] has [tex]n[/tex] as its argument and [tex]n+1[/tex] as its base.
This creates a chain where each [tex]\log n[/tex] in the numerator is cancelled by [tex]\log n[/tex] in the denominator of the next term.

So, let's see how this simplification works:

[tex]\frac{\log 2}{\log 3} \cdot \frac{\log 3}{\log 4} \cdot \frac{\log 4}{\log 5} \cdots \frac{\log 15}{\log 16}[/tex]

All intermediate terms cancel out, leaving only:

[tex]\frac{\log 2}{\log 16}[/tex]

Using the change of base formula:

[tex]\log_{16} 2[/tex] can be expressed as [tex]\frac{1}{4}[/tex] because [tex]16 = 2^4[/tex].

Thus, the entire product simplifies to:

[tex]\boxed{\log_{16} 2} = \boxed{\frac{1}{4}}[/tex]

Therefore, the value of the given expression is [tex]\frac{1}{4}[/tex].