Logarithms' challenger #1

1. First we convert all of the logs into the same base using $\log_{a}{b} = \frac{\log_{c}{b}}{\log_{c}{a}}$

We'll convert to a base of $3$ since that's the first base used in the question.

$\log_{3}{4} \cdot \frac{\log_{3}{5}}{\log_{3}{4}} \cdot \frac{\log_{3}{6}}{\log_{3}{5}} \cdot \frac{\log_{3}{7}}{\log_{3}{6}} \cdot \frac{\log_{3}{8}}{\log_{3}{7}} \cdot \frac{\log_{3}{9}}{\log_{3}{8}} = ?$

1. Simplifying this down gives us:

$\log_{3}{9} = ?$

1. We can then find the answer by simplifying further.

$\log_{3}{3^{2}} = ?$

$2\log_{3}{3} = ?$

$2 \cdot 1 = ?$

$? = 2$

The answer is $\boxed{2}$

Using the base changing rule (here I will use natural logarithm as the new base), we will get that $log_{a}b = \frac{ln(b)}{ln(a)}$ With this it is possible to see that the remaining terms will be $\frac{ln(9)}{ln(3)}=\frac{2ln(3)}{ln(3)} = 2$