<s>[INST] No. f ordered pair $(a,r)$ in Logarithmic equation. If $a_{1},a_{2},a_{3},.............$ be a Geometric Progression, Where $a_{1} = a$ and common ratio $r$ are
positive integers. If $\displaystyle \log_{8}(a_{1})+\log_{8}(a_{12})+\log_{8}(a_{3})+....+\log_{8}(a_{12}) = 2014\;,$ Then the no. of
ordered pair $(a,r)$ are 
$\bf{My\; Try::}$ Given $a_{1},a_{2},a_{3},..........a_{12}$ are in Geometric Progression, Then 
$a_{1}=a\;\;,a_{2}=ar\;\;,a_{3}=ar^2\;\;,.................,a_{12}=ar^{11}$
Then We can Write above equation as $\displaystyle \log_{8}\left(a_{1}\cdot a_{2}\cdot a_{3}\cdot .........\cdot a_{11}\right) = 2014$
So Equation $a_{1}\cdot a_{2}\cdot a_{3}\cdot .........\cdot a_{11}=8^{2014}$
So equation $\displaystyle a\cdot ar \cdot ar^2 \cdot ........\cdot ar^{11}=8^{2014}$
So we get $\displaystyle a^{12}\cdot r^{66}=8^{2014}\;,$ Where $a,r\in \mathbb{Z^{+}}$
Now How can I found Total no. of Positive Integer ordered pair $(a,r)$
Help Required, Thanks
 [/INST] $$a^{12} \cdot r^{66} = 2^{3 \cdot 2014}$$
For positive integer numbers, we can simply divide the powers with $6$:
$$(a^2 \cdot r^{11})^6=(2^{1012})^6$$
$$a^2\cdot r^{11}=2^{1012}$$
Set $a=2^x$ and $r=2^y$ and the question turns to # of $(x,y)$ pairs:
$$2^{2x}\cdot 2^{11y}=2^{1012}$$
And then:
$$2x+11y=1012$$
The last stage:
First pair is $(506, 0)$ If we increment $y$ by $2$, the $x$ will decrement by $11$ so:
$$(506, 0), (495, 2), (484, 4), (473, 6), ... (0, 92)$$
The number of pairs is
$$\frac{506 - 0}{11} + 1 = 47$$
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