The maclaurin series is a special case of the Taylor expansion with #x_0 = 0#.
Taylor Expansion:
#f(x) = f(x_0) + (f'(x_0))/(1!)(x-x_0) + (f''(x_0))/(2!)(x-x_0)^2 + (f'''(x_0))/(3!)(x-x_0)^3+...#
So Maclaurin series will be:
#f(x) = f(0) + (f'(0))/(1!)x + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + ...#
We have #f(x) = e^(7x)ln((1-x)/3)#
I'm only going to compute up to the second derivative, ie quadratic terms of x because this function will just get so messy it'll be easy to make a simple mistake. If you need more either work them out or use one of the various online tools.
Using product rule and chain rule:
#f'(x) = d/(dx)(e^(7x))ln((1-x)/3) + e^(7x)d/(dx)(ln((1-x)/3)))#
#f'(x) = 7e^(7x)ln((1-x)/3) + e^(7x) * (3)/(1-x) * (-1/3)#
#f'(x) = e^(7x)(7ln((1-x)/3) + 1/(x-1))#
#f''(x) = d/(dx)(e^(7x))(7ln((1-x)/3) + 1/(x-1)) + e^(7x)d/(dx)(7ln((1-x)/3) + 1/(x-1)))#
#f''(x) = 7e^(7x)(7ln((1-x)/3) + 1/(x-1)) + e^(7x)(7/(x-1) - 1/(x-1)^2)#
#f''(x) = e^(7x)[49ln((1-x)/3) + 14/(x-1) - 1/(x-1)^2]#
So now that we have these, calculate #f(0), f'(0) and f''(0)#
#f(0) = e^0ln(1/3) = ln(1/3)#
#f'(0) = e^0[7ln(1/3) - 1] = 7ln(1/3) -1#
#f''(0) = e^0[49ln(1/3) - 14 - 1] = 49ln(1/3) - 15#
Hence:
#f(x) = ln(1/3) + (7ln(1/3) - 1)x + (49ln(1/3) - 15)/(2)x^2 +...#
