Series of logarithms

We can write the given equation as:

$\frac{\log a}{\log 3} + \frac{\log a}{\frac{1}{2}\log 3} + \frac{\log a}{\frac{1}{3}\log 3} + \frac{\log a}{\frac{1}{4}\log 3} + \ldots + \frac{\log a}{\frac{1}{10}\log 3} = 110$

then factor out $\frac{\log a}{\log 3}$

$\frac{\log a}{\log 3}(1 + \frac{1}{\frac{1}{2}} + \frac{1}{\frac{1}{3}} + \frac{1}{\frac{1}{4}} + \ldots + \frac{1}{\frac{1}{10}}) = 110$

$\Rightarrow \frac{\log a}{\log 3}(1 + 2 + 3 + 4 + \ldots + 10) = 110$

$\Rightarrow \frac{\log a}{\log 3}(55) = 110$

$\Rightarrow \frac{\log a}{\log 3} = \frac{110}{55} = 2$

$\Rightarrow \log a = 2\log 3 = \log 3^2 = \log 9$

which implies $a = \boxed{9}$

Hi Shabarish, I enjoyed this problem!!!! :)