# graphing problem

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• 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

In addition:

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?

please help Idon't understand the problem. thanks!
• 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:

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

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:

In addition:

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?

please help Idon't understand the problem. thanks!
1. Graph $f(x)=\ln(x)$

2. f(x-3) means the graph of f is translated by 3 units to the right.

3. -f(x-3) means the translated graph is refelcted about the x-axis.

4. -f(x-3) + 2 means the translated and reflected graph is translated by 2 units upwards.

5. The x-coordinate of the point of intersection is the solution to the equation above:

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