Let $a_1,a_2,...$ be a geometric progression, where $a_1=a$ and common ratio is r 
a and r are positive integers. If $$\log_8a_1+\log_8 a_2....\log_8 a_{12}=2014$$, then find number of orderered pairs (a,r)

The expression is 
$$\log_8(a_1a_2a_3....a_{12})=2014$$
$$\log_8(a^{12}r^{66})=2014$$
$$6\log_8(a^2r^{11})=2014$$
How should I solve further?
 A: I would go back to your previous step and raise $8$ to the power of each side, then take the base $2$ log getting
$$a^{12}r^{66}=8^{2014}=2^{6042}\\12\log_2 a + 66 \log_2 r=6042$$
Note that $\log_2 a$ and $\log_2 r$ must be integers, so let them be $A,R$
$$12A+66R=6042\\
2A+11R=1007$$
One solution is $A=498,R=1$ and you can remove $11\ A$s and add $2\ R$s as many times as you like until $A$ goes negative.  That gives $46$ solutions.
A: We have 
\begin{eqnarray*}
\log_2(a^2r^{11}) =1007.
\end{eqnarray*}
$a$ and $r$ will need to be powers of $2$, so let $a=2^x$ and $r=2^y$ and 
\begin{eqnarray*}
2x+11y =1007.
\end{eqnarray*}
The positive integer solutions can be parametrised by $x=3+11t$ and $y=91-2t$ where $t=0,1,\cdots,45$. So there are $\color{red}{46}$ solutions.
A: We have 
$\log_2(a^2r^{11}) =1007\implies$ $2^{1007}=a^2r^{11}$
Hence $(a,r)$ are powers of $2$. Let $a=2^x$ and $r=2^y$
Therefore, $2^{1007}=2^{2x}2^{11y} =2^{2x+11y}$
$2x+11y=1007$
Note here that RHS is odd, therefore LHS has to be odd.
$2x$ is even. Even+Odd=Odd. Therefore $11y$ has to be odd. Therefore $y$ cannot be of the form $2k$ because it cannot be even. So start by taking $y=1,3,5...$ If you divide $1007$ by $11$ you get $91$ as the quotient. So the maximum possible value of $y$ is $91$. The corresponding value of $x$ in this case is $3$. 
The number of solutions is same as the number of possible values of $y$ which is the number of odd numbers from$1$ to $91$, i.e., $46$. The pairs $(a,r)$ are $(2^{996},2^{11})$...$(2^6,2^{1001})$.
