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To compute $\log_{8^a} 4^b$ in terms of $a$ and $b$, we'll utilize the properties of logarithms.

First, let's express $8^a$ and $4^b$ in terms of a common base. Since both 8 and 4 can be represented as powers of 2, we rewrite them as follows:

$$8^a = (2^3)^a = 2^{3a}$$

$$4^b = (2^2)^b = 2^{2b}$$

Now, our expression becomes:

$$\log_{8^a} 4^b = \log_{2^{3a}} 2^{2b}$$

Using the property of logarithms that $\log_a b^n = n \cdot \log_a b$, we rewrite this as:

$$= 2b \cdot \log_{2} 2^{3a}$$

Since $\log_{2} 2^{3a} = 3a$ (as $\log_{a} a = 1$), our expression simplifies to:

$$= 2b \cdot 3a$$

Thus, $\log_{8^a} 4^b$ in terms of $a$ and $b$ is $6ab$.