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

Quizplus · Accepted Answer

The expression can be rewritten using the power rule of logarithms: $-2 \log_b x = \log_b (x^{-2})$, $-6 \log_b y = \log_b (y^{-6})$, and $\frac{1}{7} \log_b z = \log_b (z^{1/7})$. Combining these gives $\log_b \left(\frac{z^{1/7}}{x^2 y^6}\right)$.

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 $- 2 \log _ { b } x - 6 \log _ { b } y + \frac { 1 } { 7 } \log _ { b } z$