Hello
How would one get from
$\displaystyle
\frac{x(1-\frac{4}{x})-(x-4\ln(x))}{1-\frac{4}{x}}
$
To
$\displaystyle
\frac{4x(1-\ln(x))}{4-x}
$
Hello, macca101!
Did you try any algebra on it?
How would one get from: $\displaystyle \frac{x(1-\frac{4}{x})-(x-4\ln(x))}{1-\frac{4}{x}}$ to $\displaystyle \frac{4x(1-\ln(x))}{4-x}$
Did you simplify the numerator?
. . $\displaystyle x\left[1 - \frac{4}{x}\right] - \left[x - 4\ln(x)\right] \;=$ $\displaystyle \;x - 4 - x + 4\ln(x) \;=\;-4 + 4\ln(x)$
So we have: .$\displaystyle \frac{-4 + 4\ln(x)}{1 - \frac{4}{x}} $
Multiply top and bottom by $\displaystyle x:\;\;\frac{x[-4 + 4\ln(x)]}{x - 4} $
Factor: .$\displaystyle \frac{-4x[1 - \ln(x)]}{-(4 - x)} \;= \;\frac{4x[1 - \ln(x)]}{4 - x} $