# How do you graph f(X)=ln(2x-6)?

Mar 31, 2016

Find the key points of a logarithm function:

$\left({x}_{1} , 0\right)$

$\left({x}_{2} , 1\right)$

$\ln \left(g \left(x\right)\right) \to g \left(x\right) = 0$ (vertical asymptote)

Keep in mind that:

$\ln \left(x\right) \to$increasing and concave
$\ln \left(- x\right) \to$decreasing and concave

#### Explanation:

$f \left(x\right) = 0$

$\ln \left(2 x - 6\right) = 0$

$\ln \left(2 x - 6\right) = \ln 1$

$\ln x$ is $1 - 1$

$2 x - 6 = 1$

$x = \frac{7}{2}$

• So you have one point $\left(x , y\right) = \left(\frac{7}{2} , 0\right) = \left(3.5 , 0\right)$

$f \left(x\right) = 1$

$\ln \left(2 x - 6\right) = 1$

$\ln \left(2 x - 6\right) = \ln e$

$\ln x$ is $1 - 1$

$2 x - 6 = e$

$x = 3 + \frac{e}{2} \cong 4.36$

• So you have a second point $\left(x , y\right) = \left(1 , 4.36\right)$

Now to find the vertical line that $f \left(x\right)$ never touches, but tends to, because of its logarithmic nature. This is when we try to estimate $\ln 0$ so:

$\ln \left(2 x - 6\right)$

$2 x - 6 = 0$

$x = 3$

• Vertical asymptote for $x = 3$
• Finally, since the function is logarithmic, it will be increasing and concave .

Therefore, the function will:

• Increase but curve downwards.
• Pass through $\left(3.5 , 0\right)$ and $\left(1 , 4.36\right)$
• Tend to touch $x = 3$

Here is the graph:

graph{ln(2x-6) [0.989, 6.464, -1.215, 1.523]}