+218 Suppose a does not equal 0. Compute log_{8^a} 4^b in terms of a and b.
+1588 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 \( 6ab \).
