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Vlad Balmos · Answer 1 · 2023-05-17T10:31:35+0000

Solution (b):

Given the logarithm;

$4\log_6x+5\log_6y$

Following the addition property and power property of logarithm written as;

$\begin{gathered} \log_ca+\log_cb=\log_c(ab) \\ \\ b\log_ca=\log_c(a)^b \end{gathered}$

Thus;

$\begin{gathered} 4\operatorname{\log}_6x+5\operatorname{\log}_6y=\log_6(x)^4+\log_6(y)^5 \\ \\ 4\operatorname{\log}_6x+5\operatorname{\log}_6y=\log_6(x^4y^5) \end{gathered}$

ANSWER:

$\begin{equation*} \log_6(x^4y^5) \end{equation*}$

Solution (d):

Given;

$2\log_3x+(1)/(3)\log_3x-2\log_3(x+1)$
$\begin{gathered} 2\operatorname{\log}_3x+(1)/(3)\operatorname{\log}_3x-2\operatorname{\log}_3(x+1)=\log_3(x)^2+\log_3(x)^{(1)/(3)}-\log_3(x+1)^2 \\ \\ 2\operatorname{\log}_3x+(1)/(3)\operatorname{\log}_3x-2\operatorname{\log}_3(x+1)=\log_3(\frac{x^2* x^{(1)/(3)}}{(x+1)^2}) \\ \\ 2\operatorname{\log}_3x+(1)/(3)\operatorname{\log}_3x-2\operatorname{\log}_3(x+1)=\log_3(\frac{x^{2+(1)/(3)}}{x^2+2x+1}) \\ \\ 2\operatorname{\log}_3x+(1)/(3)\operatorname{\log}_3x-2\operatorname{\log}_3(x+1)=\log_3(\frac{x^{(7)/(3)}}{x^2+2x+1}) \end{gathered}$

ANSWER:

$\begin{equation*} \log_3(\frac{x^{(7)/(3)}}{x^2+2x+1}) \end{equation*}$

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