How do you find the maclaurin series expansion of #e^(7x)ln((1-x)/3)#?

Taylor Expansion:

So Maclaurin series will be:

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:

Hence:

To find the Maclaurin series expansion of $e^{7x} \ln\left(\frac{1-x}{3}\right)$, we use the formula for the Maclaurin series expansion of a composite function. The formula states that if $f(x)$ and $g(x)$ have Maclaurin series expansions, then the Maclaurin series expansion of $f(g(x))$ is obtained by substituting the series expansion of $g(x)$ into $f(x)$, term by term.

Here, $f(x) = e^x$ and $g(x) = 7x \ln\left(\frac{1-x}{3}\right)$.

The Maclaurin series expansion of $e^x$ is $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$.

For $g(x) = 7x \ln\left(\frac{1-x}{3}\right)$, we need to find its Maclaurin series expansion. We'll use the properties of logarithms and the geometric series to expand $\ln\left(\frac{1-x}{3}\right)$.

$\ln\left(\frac{1-x}{3}\right) = \ln(1-x) - \ln(3)$

$= -\left(x + \frac{x^2}{2} + \frac{x^3}{3} + \ldots\right) - \ln(3)$

Now, multiply by $7x$:

$7x \ln\left(\frac{1-x}{3}\right) = -7x^2 - \frac{7x^3}{2} - \frac{7x^4}{3} - \ldots - 7x\ln(3)$

Now, substitute this series expansion into the Maclaurin series expansion of $e^{7x}$:

$e^{7x} \ln\left(\frac{1-x}{3}\right) = e^{7x}(-7x^2 - \frac{7x^3}{2} - \frac{7x^4}{3} - \ldots - 7x\ln(3))$

This gives us the Maclaurin series expansion of $e^{7x} \ln\left(\frac{1-x}{3}\right)$.