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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 \( 6ab \).

bader Mar 30, 2024