Simplifying Product of Logarithms

Question

I'm trying to simplify

$$\log_{10}11 \cdot \log_{11}12 \cdot \log_{12}13 \cdot \ldots \cdot \log_{998}999 \cdot \log_{999}1000\ .$$

I have absolutely no clue where to start in simplifying it. I know the answer is 3, but I have no clue how to get there.

Answer 1

Hint: $$\log_ab=\frac{\log_{10}b}{\log_{10}a}\ .$$

Answer 2

Use the change base formula $\log_a b=\dfrac{\log_c b}{\log_c a}$, we have $\log_{10} 11*\log_{11} 12*\dots \log_{999} 1000=\left( \dfrac{\log_{10} 11}{\log_{10} 10}\right) *\left( \dfrac{\log_{10} 12}{\log_{10} 11}\right)*\left( \dfrac{\log_{10} 13}{\log_{10} 12}\right)*\dots*\left( \dfrac{\log_{10} 999}{\log_{10} 998}\right)*\left( \dfrac{\log_{10} 1000}{\log_{10} 999}\right)=\left( \dfrac{\log_{10} 1000}{\log_{10} 10}\right)=\log_{10}1000=3 $

Answer 3

$$ \log_{a} b \times \log_{b} c = \log_{a} c$$

$$\log_{10} 11\times \log_{11} 12 \times \dots \times\log_{999} 1000=\log_{10} 1000=3$$

Answer 4

In a more general manner, you are looking for $$A=\prod_{i=10}^b\log_{i}(i+1)$$ Back to natural logarithms, this gives $$A=\prod_{i=10}^b\frac{\log (i+1)}{\log (i)}$$ and you see that the numerator of the $n^{th}$ term is identical to the denominator of the $(n+1)^{th}$ term; so, a lot of simplifications are possible. At the end $$A=\frac{\log (b+1)}{\log (10)}$$