Assume that x, y, and z are positive numbers.Use the properties of logarithms to write the expression $- 2 \log _ { b } x - 6 \log _ { b } y + \frac { 1 } { 7 } \log _ { b } z$ as the logarithm of one quantity.
A) $\log _ { b } \frac { z ^ { 1 / 7 } } { x ^ { 6 } y ^ { 2 } }$
B) $\log _ { b } \frac { z ^ { 1 / 6 } } { x ^ { 2 } y ^ { 7 } }$
C) $\log _ { b } \frac { z ^ { 1 / 2 } } { x ^ { 7 } y ^ { 6 } }$
D) $\log _ { b } \frac { x ^ { 1 / 7 } } { y ^ { 2 } z ^ { 6 } }$
E) $\log _ { b } \frac { z ^ { 1 / 7 } } { x ^ { 2 } y ^ { 6 } }$

Question

Assume that x, y, and z are positive numbers.Use the properties of logarithms to write the expression $- 2 \log _ { b } x - 6 \log _ { b } y + \frac { 1 } { 7 } \log _ { b } z$ as the logarithm of one quantity. 
A) $\log _ { b } \frac { z ^ { 1 / 7 } } { x ^ { 6 } y ^ { 2 } }$
B) $\log _ { b } \frac { z ^ { 1 / 6 } } { x ^ { 2 } y ^ { 7 } }$
C) $\log _ { b } \frac { z ^ { 1 / 2 } } { x ^ { 7 } y ^ { 6 } }$
D) $\log _ { b } \frac { x ^ { 1 / 7 } } { y ^ { 2 } z ^ { 6 } }$
E) $\log _ { b } \frac { z ^ { 1 / 7 } } { x ^ { 2 } y ^ { 6 } }$

Quizplus · Accepted Answer

The expression can be simplified using the properties of logarithms: the coefficient in front of a logarithm can be moved to the exponent of the argument, and subtraction between logarithms translates to division inside the logarithm, while addition translates to multiplication. Thus, $- 2 \log _ { b } x - 6 \log _ { b } y + \frac { 1 } { 7 } \log _ { b } z$ simplifies to $\log _ { b } \frac { z ^ { 1 / 7 } } { x ^ { 2 } y ^ { 6 } }$.

Assume That X, Y, and Z Are Positive Numbers −2log⁡bx−6log⁡by+17log⁡bz- 2 \log _ { b } x - 6 \log _ { b } y + \frac { 1 } { 7 } \log _ { b } z−2logb​x−6logb​y+71​logb​z

Assume That X, Y, and Z Are Positive Numbers $- 2 \log _ { b } x - 6 \log _ { b } y + \frac { 1 } { 7 } \log _ { b } z$