which expression is equivalent to $log_{c}\frac{x^{2}-1}{5x}$?\n$log_{c}x^{2}-log_{c}5x - 1$\n$2log_{c}x-(log…

which expression is equivalent to $log_{c}\frac{x^{2}-1}{5x}$?\n$log_{c}x^{2}-log_{c}5x - 1$\n$2log_{c}x-(log_{c}5+log_{c}x)-1$\n$log_{c}(x^{2}-1)-(log_{c}5+log_{c}x)$\n$2log_{c}x-log_{c}1-log_{c}5+log_{c}x$

which expression is equivalent to $log_{c}\frac{x^{2}-1}{5x}$?\n$log_{c}x^{2}-log_{c}5x - 1$\n$2log_{c}x-(log_{c}5+log_{c}x)-1$\n$log_{c}(x^{2}-1)-(log_{c}5+log_{c}x)$\n$2log_{c}x-log_{c}1-log_{c}5+log_{c}x$

Answer

Explanation:

Step1: Apply logarithm quotient - rule

The quotient - rule of logarithms states that $\log_a\frac{M}{N}=\log_aM-\log_aN$. For $\log_c\frac{x^{2}-1}{5x}$, we have $\log_c(x^{2}-1)-\log_c(5x)$.

Step2: Apply logarithm product - rule

The product - rule of logarithms states that $\log_a(MN)=\log_aM+\log_aN$. Since $5x = 5\times x$, then $\log_c(5x)=\log_c5+\log_cx$. So, $\log_c\frac{x^{2}-1}{5x}=\log_c(x^{2}-1)-(\log_c5+\log_cx)$.

Answer:

$\log_c(x^{2}-1)-(\log_c5+\log_cx)$