Question

How would you write the following expression as a single term? 3[2 ln(x-1) - lnx] + ln (x+1)

Answer 1

Apply the rule: $n ln x = ln x^{n}$

$3[2 ln(x-1) - lnx] + ln(x+1)=3[ln(x-1)^{2} - lnx ] + ln(x+1)$

Apply the rule : $log a - log b = log \frac{a}{b}$

$3[2 ln(x-1) - lnx] + ln(x+1)=3ln\frac{(x-1)^{2} }{x} + ln(x+1)$

Apply the rule: $n ln x = ln x^{n}$

$3[ln (x-1)^{2} -ln x]+ln (x+1)= ln \frac{(x-1)^{6} }{x^{3} } +log(x+1)$

Finally, apply the rule: log a + log b = log ab

$3[ln(x-1)^{2} -ln x]+log(x+1)=ln\frac{(x-1)^{6}(x+1) }{x^{3} }$