Question

Express in the form y=f(x):
In(4-y)=In(x-2)+In(8-x)-x

Answer 1

\[e^{\ln(4-y)}=e ^{\ln(x-2)(8-x)-x}\]
\[4-y=\frac{(x-2)(8-x)}{e ^{x}} \]
\[y=-\frac{(x-2)(8-x)}{e ^{x}} + 4\]
\[y=\frac{x^2-10x+4e^x+16}{e^x}\]

Answer 2

thax loads anhhuyalex,i understand the first three steps but not the last one-may you please elaborate how we got 4e^x or which rule are we using?

Answer 3

the last step is
\[y=-\frac{-x^2+10x-16}{e^x}+4=\frac{x^2-10x+16}{e^x}+4=\frac{x^2-10x+16+4e^x}{e^x}\]
because 4e^x/e^x is essentially 4. I'm only using algebra to manipulate the last bit.