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Zepler [3.9K]
4 years ago
8

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

Mathematics
1 answer:
Elan Coil [88]4 years ago
3 0

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} }

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