# Math Help - graphing problem

1. ## graphing problem

i have this equation:

solve for x:
ln(x)+ln(x-3)=2

x=0, x=3

graph f(x)=ln(x) and -f(x-3)+2 by using transformations. Explain the relationship between the solutions to the original equation and the points of intersection on your graph.

How do you graph -f(x-3)+2?

2. Originally Posted by Anemori
i have this equation:

solve for x:
ln(x)+ln(x-3)=2

x>0 and x>3 <<<<<<<<< typo
Use the laws of logarithms:

$\ln(x)+\ln(x-3)=2~\implies~\ln(x(x-3))=\ln(e^2)$

Now use the logarithms as exponents to the base e. You'll get a quadratic equation. Keep in mind that this equation is only valid for x > 3

graph f(x)=ln(x) and -f(x-3)+2 by using transformations. Explain the relationship between the solutions to the original equation and the points of intersection on your graph.

How do you graph -f(x-3)+2?

1. Graph $f(x)=\ln(x)$
$\ln(x)+\ln(x-3)=2~\implies~\underbrace{\ln(x)}_{f(x)}=\underbra ce{-\ln(x-3)+2}_{-f(x-3)+2}$