$$ \log _{2} x=6 $$ y $$ \log _{3} y=1 \Rightarrow x+y $$

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$$ x + y = 67 $$ To solve the equations $$ \log_{2} x = 6 $$ and $$ \log_{3} y = 1 $$, we will first convert these logarithmic equations into their exponential forms. 1. For the first equation $$ \log_{2} x = 6 $$: $$ x = 2^6 = 64 $$ 2. For the second equation $$ \log_{3} y = 1 $$: $$ y = 3^1 = 3 $$ Now that we have the values of $$ x $$ and $$ y $$, we can find $$ x + y $$: $$ x + y = 64 + 3 = 67 $$ Thus, the final answer is: $$ \boxed{67} $$

\( \log _{2} x=6 \) y \( \log _{3} y=1 \Rightarrow x+y \)

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\( \log _{2} x=6 \) y \( \log _{3} y=1 \Rightarrow x+y \)

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