Rewrite the expression as the sum and/or difference of logarithms, without using exponents. Simplify if possible.
-$$\log _ { 2 } \frac { \sqrt [ 5 ] { m } \sqrt [ 9 ] { n } } { k ^ { 2 } }$$
A) $5 \log _ { 2 } m + 9 \log _ { 2 } n - 2 \log _ { 2 } k$
B) $\frac { 5 } { 2 } \log _ { 2 } \mathrm {~m} + \frac { 9 } { 2 } \log _ { 2 } \mathrm { n } - \frac { 1 } { 2 } \log _ { 2 } \mathrm { k }$
C) $\frac { 1 } { 5 } \log _ { 2 } \mathrm {~m} \cdot \frac { 1 } { 9 } \log _ { 2 } \mathrm { n } \div 2 \log _ { 2 } \mathrm { k }$
D) $\frac { 1 } { 5 } \log _ { 2 } \mathrm {~m} + \frac { 1 } { 9 } \log _ { 2 } \mathrm { n } - 2 \log _ { 2 } \mathrm { k }$

Question

Quizplus · Accepted Answer

The expression can be rewritten using the properties of logarithms: the quotient rule $\log_b \frac{x}{y} = \log_b x - \log_b y$, the product rule $\log_b (xy) = \log_b x + \log_b y$, and the power rule $\log_b x^n = n \log_b x$. The square root and other roots can be expressed as fractional exponents, leading to the correct simplification.

Rewrite the Expression as the Sum And/or Difference of Logarithms log⁡2m5n9k2\log _ { 2 } \frac { \sqrt [ 5 ] { m } \sqrt [ 9 ] { n } } { k ^ { 2 } }log2​k25m​9n​​

Rewrite the Expression as the Sum And/or Difference of Logarithms $\log _ { 2 } \frac { \sqrt [ 5 ] { m } \sqrt [ 9 ] { n } } { k ^ { 2 } }$