$\\log a = 4$ $\\log b = 9$ $\\log c = 4$\n$\\log \\frac{b^7}{a^3c^2}$\nanswer attempt 1 out of 2
$\\log a = 4$ $\\log b = 9$ $\\log c = 4$\n$\\log \\frac{b^7}{a^3c^2}$\nanswer attempt 1 out of 2
Answer
Explanation:
Step1: Apply log quotient rule
$\log\frac{b^7}{a^3c^2} = \log b^7 - \log(a^3c^2)$
Step2: Apply log product rule
$\log b^7 - \log(a^3c^2) = \log b^7 - (\log a^3 + \log c^2)$
Step3: Simplify the expression
$\log b^7 - \log a^3 - \log c^2$
Step4: Apply log power rule
$7\log b - 3\log a - 2\log c$
Step5: Substitute given values
$7(9) - 3(4) - 2(4)$
Step6: Calculate the result
$63 - 12 - 8 = 43$
Answer:
43