If \( \log 12 \approx 1.0792 \) and \( \log 3 \approx 0.4771 \), find \( \log 36 \) to four decimal places. (1 point) \( \log 36 \approx \) Check answer Remaining Attempts : 3

To find \( \log 36 \), you can use the fact that \( 36 = 6^2 \). Therefore, \[ \log 36 = \log(6^2) = 2 \log 6. \] Next, we can express \( \log 6 \) in terms of \( \log 2 \) and \( \log 3 \). Since \( 6 = 2 \times 3 \), we have: \[ \log 6 = \log 2 + \log 3. \] However, we first need to find \( \log 2 \). We can use the approximation of \( \log 12 \): Knowing that \( 12 = 4 \times 3 = 2^2 \times 3 \), \[ \log 12 = \log(2^2) + \log(3) = 2 \log 2 + \log 3. \] Substituting the known values: \[ 1.0792 = 2 \log 2 + 0.4771. \] Now, rearranging to find \( \log 2 \): \[ 2 \log 2 = 1.0792 - 0.4771 = 0.6021, \] \[ \log 2 = \frac{0.6021}{2} = 0.30105. \] Now plug \( \log 2 \) and \( \log 3 \) back into the equation for \( \log 6 \): \[ \log 6 = \log 2 + \log 3 = 0.30105 + 0.4771 = 0.77815. \] Finally, we compute \( \log 36 \): \[ \log 36 = 2 \log 6 = 2 \times 0.77815 = 1.5563. \] So, \( \log 36 \approx 1.5563 \).