Apply the rule: 
![3[2 ln(x-1) - lnx] + ln(x+1)=3[ln(x-1)^{2} - lnx ] + ln(x+1)](https://tex.z-dn.net/?f=3%5B2%20ln%28x-1%29%20-%20lnx%5D%20%2B%20ln%28x%2B1%29%3D3%5Bln%28x-1%29%5E%7B2%7D%20-%20lnx%20%5D%20%2B%20ln%28x%2B1%29)
Apply the rule : 
![3[2 ln(x-1) - lnx] + ln(x+1)=3ln\frac{(x-1)^{2} }{x} + ln(x+1)](https://tex.z-dn.net/?f=3%5B2%20ln%28x-1%29%20-%20lnx%5D%20%2B%20ln%28x%2B1%29%3D3ln%5Cfrac%7B%28x-1%29%5E%7B2%7D%20%7D%7Bx%7D%20%2B%20ln%28x%2B1%29)
Apply the rule: 
![3[ln (x-1)^{2} -ln x]+ln (x+1)= ln \frac{(x-1)^{6} }{x^{3} } +log(x+1)](https://tex.z-dn.net/?f=3%5Bln%20%28x-1%29%5E%7B2%7D%20-ln%20x%5D%2Bln%20%28x%2B1%29%3D%20ln%20%5Cfrac%7B%28x-1%29%5E%7B6%7D%20%7D%7Bx%5E%7B3%7D%20%7D%20%2Blog%28x%2B1%29)
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} }](https://tex.z-dn.net/?f=3%5Bln%28x-1%29%5E%7B2%7D%20-ln%20x%5D%2Blog%28x%2B1%29%3Dln%5Cfrac%7B%28x-1%29%5E%7B6%7D%28x%2B1%29%20%7D%7Bx%5E%7B3%7D%20%7D)
Answer: The value of m is 29.
Step-by-step explanation:
Given that, One term of
is
...(i)
We know that that (r+1)th term in
is given by :-
...(ii)
On comparing (i) with (ii) , we get

i.e.

Hence, the value of m is 29.
<u>Part</u><u> </u><u>1</u> 6(1.20)+4(1.05)=$11.40
<u>Part</u><u> </u><u>2</u> 1.20x+1.05y
31/36 is the correct answer