How do you express $\log _ { 6} 11z$ as the sum or difference of logarithms?

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Answer 1 · 2017-10-11T17:03:12

$log_6(11) + log_6(z)$

There exists a property of logarithms that states that:

$log_x(a*b) = log_x(a) + log_x(b)$

With this in mind, we can rewrite $log_6(11*z)$ as:

$log_6(11) + log_6(z)$

How do you express \log _ { 6} 11z\log _ { 6} 11z as the sum or difference of logarithms?