Log Question, Help ASAP

Question

Let $P = \log_a b,$ where

$$P = \log_2 3 \cdot \log_3 4 \cdot \log_4 5 \cdots \log_{2008} 2009$$ and $a$ and $b$ are relatively prime positive integers. Find the smallest possible value of $a+b.$

Answer 1

To understand this you need to know the base change rule for logarithms. The base change rule is $$\log_a b =\frac{\log_x b}{\log_x a}$$

This further tells us that $P$ can be simplified into one giant fraction being equal to $$\frac{\log_x 3 \cdot\ldots\cdot \log_x 2009}{\log_x 2 \cdot \ldots \cdot \log_x 2008}$$

Here we can see there are a lot of common terms between the numerator and denominator that can be cancelled from $\log_x 3$ to $\log_x 2008$.

This leaves us with a much simplified fraction that is equal to $$\frac{\log_x 2009}{\log_x 2}$$

The exact opposite of he base change rule for logarithms can now be applied to change the above fraction into $\log_2 2009$. This is our desired result as we see $a=2,b=2009$, which are relatively prime. Thus showing that $a+b=2011$

Answer 2

By change of base and cancelling like terms, we have $P = \frac{\log 3}{\log 2} \times \frac{\log 4}{\log 3} \times \frac{\log 5}{\log 4} \cdots \frac{\log 2009}{\log 2008} = \frac{\log 2009}{\log 2} = \log_2 2009$.