graphing problem

• May 18th 2010, 11:52 PM
Anemori
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?

• May 19th 2010, 05:37 AM
earboth
Quote:

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:

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

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

Quote:

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