2015 Mock AIME I Problem 2: Logarithm System

$Let\quad \log _{ 3 }{ x } =X\quad \& \quad \log _{ 3 }{ y } =Y\quad then,\\ \\ \cfrac { 20 }{ 13 } =\log _{ x }{ 3y } =\cfrac { 1+Y }{ X } \\ \\ \Rightarrow 20X-13Y-13=0\quad \quad \quad ....(i)\\ \\ \& \cfrac { 2 }{ 3 } =\log _{ 3x }{ y } =\cfrac { Y }{ 1+X } \\ \\ \Rightarrow 2X-3Y+2=0\quad \quad \quad \quad \quad ...(ii)\\ \\ solving\quad these,\\ \\ X=\cfrac { 65 }{ 34 } ,Y=\cfrac { 66 }{ 34 } \Rightarrow \log _{ 3x }{ 3y } =\cfrac { 1+Y }{ 1+X } =\cfrac { 100 }{ 99 } \\ \\ \Rightarrow a+b=100+99=199$

$\log _{3x} { 3y } = \log _{3x} { 3 } + \log _{3x} { y }$

Since $\log _{3x} { y } = \frac {2} {3}$ is given, we only need to find $\log _{3x} { 3 }$ .

From:

$\log _{x} { 3y } = \frac {20}{13} \quad \Rightarrow 3y = x ^ { \frac {20}{13} }$

$\log _{3x} { y } = \frac {2}{3} \quad \Rightarrow y = (3x) ^ { \frac {2}{3} } \quad \Rightarrow 3y = 3 (3x) ^ { \frac {2}{3} }$

$\Rightarrow 3y = x ^ { \frac {20}{13} } = 3 (3x) ^ { \frac {2}{3} }$

$\Rightarrow (3x) ^ { \frac {20}{13} } = 3 ^ { 1+ \frac {20}{13} } (3x) ^ { \frac {2}{3} } \quad \Rightarrow (3x) ^ {\frac {20}{13} - \frac {2}{3} } = 3 ^ {\frac {33}{13} } \quad \Rightarrow (3x) ^ {\frac {34}{39} } = 3 ^ {\frac {33}{13} }$

$\Rightarrow (3x) ^ {\frac {34}{39} \times \frac {13}{33} } = 3 \quad \Rightarrow (3x) ^ {\frac {34}{99} } = 3 \quad \Rightarrow \log_{3x} {3} = \frac {34}{99}$

$\Rightarrow \log _{3x} { 3y } = \log _{3x} { 3 } + \log _{3x} { y }$

$\quad \quad = \frac {34}{99} + \frac {2}{3} = \frac{100}{99} = \frac {a}{b}\quad \Rightarrow a+b = \boxed{199}$

Nice question.!