Write as the sum and/or difference of logarithms. Express powers as factors.
-$\ln \frac { ( 6 x ) \sqrt [ 9 ] { 1 + 3 x } } { ( x - 9 ) ^ { 7 } } , \quad x 9$
A) $6 \ln x + \frac { 1 } { 3 } \ln ( 1 + 3 x ) - 7 \ln ( x - 9 )$
B) $\ln 6 + \ln x + \frac { 1 } { 9 } \ln ( 1 + 3 x ) - 7 \ln ( x - 9 )$
C) $\ln 6 + \ln x - 9 \ln ( 1 + 3 x ) - 7 \ln ( x - 9 )$
D) $\ln 6 + \ln x + \frac { 1 } { 9 } \ln ( 1 + 3 x ) - \ln 7 - \ln ( x - 9 )$

Question

Write as the sum and/or difference of logarithms. Express powers as factors.
-$\ln \frac { ( 6 x ) \sqrt [ 9 ] { 1 + 3 x } } { ( x - 9 ) ^ { 7 } } , \quad x > 9$
A) $6 \ln x + \frac { 1 } { 3 } \ln ( 1 + 3 x ) - 7 \ln ( x - 9 )$
B) $\ln 6 + \ln x + \frac { 1 } { 9 } \ln ( 1 + 3 x ) - 7 \ln ( x - 9 )$
C) $\ln 6 + \ln x - 9 \ln ( 1 + 3 x ) - 7 \ln ( x - 9 )$
D) $\ln 6 + \ln x + \frac { 1 } { 9 } \ln ( 1 + 3 x ) - \ln 7 - \ln ( x - 9 )$

Quizplus · Accepted Answer

The given expression can be rewritten using the properties of logarithms: the logarithm of a product is the sum of the logarithms ($\ln(ab) = \ln a + \ln b$), the logarithm of a quotient is the difference of the logarithms ($\ln(a/b) = \ln a - \ln b$), and the logarithm of a power is the exponent times the logarithm ($\ln(a^b) = b \ln a$). Applying these rules, we get $\ln 6 + \ln x + \frac{1}{9}\ln(1 + 3x) - 7\ln(x - 9)$.

Write as the Sum And/or Difference of Logarithms ln⁡(6x)1+3x9(x−9)7,x>9\ln \frac { ( 6 x ) \sqrt [ 9 ] { 1 + 3 x } } { ( x - 9 ) ^ { 7 } } , \quad x > 9ln(x−9)7(6x)91+3x​​,x>9

Write as the Sum And/or Difference of Logarithms $\ln \frac { ( 6 x ) \sqrt [ 9 ] { 1 + 3 x } } { ( x - 9 ) ^ { 7 } } , \quad x > 9$