Question

Write each as a single logarithm . Assume they’re all positive numbers

Answer 1

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

ANSWER:

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