What is $a$ =?

$\displaystyle \begin{aligned} \log_a \left(10\right) + \log_a \left(10^2\right) + \log_a \left(10^3\right) + \cdots + \log_a \left(10^{10}\right) & = 110 \\ \log_a \left(10 \times 10^2 \times \cdots \times 10^{10}\right) & = 110 \\ \log_a 10^{55} & = 110 \\ \implies a^{110} & = 10^{55} \\ \implies a & = 10^{\frac{55}{110}} \\ & = \boxed{\sqrt{10}} \end{aligned}$

$\begin{aligned} \log_a (10) + \log_a (10^2) + \log_a (10^3) + \cdots + \log_a (10^{10}) & = 110 \\ \log_a (10) + 2\log_a (10) + 3\log_a (10) + \cdots +10 \log_a (10) & = 110 \\ \log_a (10) \sum_{n=1}^{10} n & = 110 \\ \frac {10(10+1)}2 \log_a (10) & = 110 \\ 55 \log_a (10) & = 110 \\ \log_a (10) & = 2 \\ \implies a^2 & = 10 \\ a & = \boxed{\sqrt{10}} \end{aligned}$

By properties of logarithms, we have $\log_a(X^Y) = Y\log_a(X)$

so $\log_a(10)+\log_a(10^2)+ \log_a(10^3)+ \dots+ \log_a(10^{10}) = 110$

$\implies \log_a(10) + 2\log_a(10) + 3\log_a(10) + \cdots + 10\log_a(10) = 110$

$55 \log_a(10) = 110$

$\log_a(10) = 2$

$\implies a^2 = 10$

$a = \boxed{\sqrt{10}}$ , $\quad$ $a \ne -\sqrt{10}$ as base of logarithm cannot be negative.

$\log_a(10)+\log_a(10^2)+ \log_a(10^3)+ \dots+ \log_a(10^{10})= 110$ $\frac{\log_{10}{10^{55}}}{\log_{10}{a}} =110$ $a = 10^{\frac{55}{110}} = \sqrt{10}$