Determine the value of $ab$ if $\log_8a+\log_4b^2=5$ and $\log_8b+\log_4a^2=7$.
This problem involves logarithms with different bases, so I want to convert them to a common base using the change of base formula: $\log_bx=\frac{\log_dx}{\log_db}$. Since 8 and 4 are both powers of 2, I will use 2 as the common base. So $\log_8a=\frac{\log_2a}{\log_28}=\frac{\log_2a}{3}$ and $\log_4b^2=\frac{\log_2b^2}{\log_24}=\frac{2\log_2b}{2}=\log_2b$. Similarly, $\log_8b=\frac{\log_2b}{3}$ and $\log_4a^2=\frac{2\log_2a}{2}=\log_2a$. Substituting these expressions into the given equations, I get $\frac{\log_2a}{3}+\log_2b=5$ and $\frac{\log_2b}{3}+\log_2a=7$. Multiplying both equations by 3, I get $\log_2a+3\log_2b=15$ and $\log_2b+3\log_2a=21$. Adding these equations, I get $4\log_2a+4\log_2b=36$. Dividing by 4, I get $\log_2a+\log_2b=9$. Using the logarithm product rule, I get $\log_2ab=9$. Exponentiating both sides with base 2, I get $ab=2^9=512$. 