Question

question
if $r(x)=log_8left(\frac{x^{3}8^{x}}{3x + 1}
ight)$, find $r(x)$.
select the correct answer below:
$r(x)=\frac{3}{x}+1-\frac{3}{3x + 1}$
$r(x)=\frac{3}{x}+1-\frac{1}{(3x + 1)ln8}$
$r(x)=\frac{3x + 1}{x^{3}8ln8}$
$r(x)=\frac{3}{xln8}+1-\frac{3}{(3x + 1)ln8}$

Explanation:

Step1: Use log - rule

$r(x)=\frac{\ln(\frac{x^{3}8^{x}}{3x + 1})}{\ln8}=\frac{1}{\ln8}(\ln(x^{3}8^{x})-\ln(3x + 1))$

Step2: Expand log - terms

$r(x)=\frac{1}{\ln8}(3\ln x+x\ln8-\ln(3x + 1))$

Step3: Differentiate

$r'(x)=\frac{1}{\ln8}(\frac{3}{x}+\ln8-\frac{3}{3x + 1})$
$=\frac{3}{x\ln8}+1-\frac{3}{(3x + 1)\ln8}$

Answer:

$r'(x)=\frac{3}{x\ln8}+1-\frac{3}{(3x + 1)\ln8}$ (last option)