log a (ab)^3 , log b a=1/7
To simplify \( \log_a (ab)^3 \), we can follow these steps:
Apply the power rule of logarithms: \[ \log_a (ab)^3 = 3 \cdot \log_a (ab) \]
Use the product rule of logarithms: \[ \log_a (ab) = \log_a a + \log_a b \] Here, \( \log_a a = 1 \) because any logarithm of a number to its own base equals 1.
So, we get: \[ \log_a (ab) = 1 + \log_a b \]
Now substituting back into the expression: \[ \log_a (ab)^3 = 3 (1 + \log_a b) = 3 + 3\log_a b \]
Next, we need to express \( \log_a b \) using the given information that \( \log_b a = \frac{1}{7} \).
From the change of base formula: \[ \log_b a = \frac{1}{\log_a b} \] Thus, we have: \[ \log_a b = \frac{1}{\log_b a} = \frac{1}{\frac{1}{7}} = 7 \]
- Now plug this back into the expression: \[ \log_a (ab)^3 = 3 + 3\cdot(7) = 3 + 21 = 24 \]
Therefore, the final answer is: \[ \log_a (ab)^3 = 24 \]
