1 Answer

Jason Rahm · Answer 1 · 2023-07-19T15:56:21+0000

Answer

Step-by-step explanation

Given:

$\log_au-\log_av+9\log_aw$

To write the above given into a single logarithm, we first apply the log rule:

$b\cdot\log_a(x)=\log_a(x^b)$

So,

$9\log_aw=\log_aw^9$

Hence,

$\operatorname{\log}_au-\operatorname{\log}_av+9\operatorname{\log}_aw=\log_au-\log_av+\log_aw^9$

Next, we apply the log rule:

$\log_a(x)-\log_a(y)=\log_a((x)/(y))$

This means that:

$\operatorname{\log}_au-\operatorname{\log}_av=\log_a((u)/(v))$

Hence,

$\log_au-\log_av+\log_aw^9=\log_a((u)/(v))+\log_aw^9$

Then, we apply the log rule:

$\log_a(x)+\log_a(y)=\log_a(xy)$

So,

$\log_a((u)/(v))+\log_aw^9=\log_a((uw^9)/(v))$

Therefore, the answer is:

$\begin{equation*} \log_a((uw^9)/(v)) \end{equation*}$

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