# 6.4 Graphs of logarithmic functions  (Page 3/8)

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Given a logarithmic function with the form $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{b}\left(x\right),$ graph the function.

1. Draw and label the vertical asymptote, $\text{\hspace{0.17em}}x=0.$
2. Plot the x- intercept, $\text{\hspace{0.17em}}\left(1,0\right).$
3. Plot the key point $\text{\hspace{0.17em}}\left(b,1\right).$
4. Draw a smooth curve through the points.
5. State the domain, $\text{\hspace{0.17em}}\left(0,\infty \right),$ the range, $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ and the vertical asymptote, $\text{\hspace{0.17em}}x=0.$

## Graphing a logarithmic function with the form f ( x ) = log b ( x ).

Graph $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{5}\left(x\right).\text{\hspace{0.17em}}$ State the domain, range, and asymptote.

Before graphing, identify the behavior and key points for the graph.

• Since $\text{\hspace{0.17em}}b=5\text{\hspace{0.17em}}$ is greater than one, we know the function is increasing. The left tail of the graph will approach the vertical asymptote $\text{\hspace{0.17em}}x=0,$ and the right tail will increase slowly without bound.
• The x -intercept is $\text{\hspace{0.17em}}\left(1,0\right).$
• The key point $\text{\hspace{0.17em}}\left(5,1\right)\text{\hspace{0.17em}}$ is on the graph.
• We draw and label the asymptote, plot and label the points, and draw a smooth curve through the points (see [link] ).

The domain is $\text{\hspace{0.17em}}\left(0,\infty \right),$ the range is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ and the vertical asymptote is $\text{\hspace{0.17em}}x=0.$

Graph $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{\frac{1}{5}}\left(x\right).\text{\hspace{0.17em}}$ State the domain, range, and asymptote.

The domain is $\text{\hspace{0.17em}}\left(0,\infty \right),$ the range is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ and the vertical asymptote is $\text{\hspace{0.17em}}x=0.$

## Graphing transformations of logarithmic functions

As we mentioned in the beginning of the section, transformations of logarithmic graphs behave similarly to those of other parent functions. We can shift, stretch, compress, and reflect the parent function $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ without loss of shape.

## Graphing a horizontal shift of f ( x ) = log b ( x )

When a constant $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ is added to the input of the parent function $\text{\hspace{0.17em}}f\left(x\right)=lo{g}_{b}\left(x\right),$ the result is a horizontal shift     $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ units in the opposite direction of the sign on $\text{\hspace{0.17em}}c.\text{\hspace{0.17em}}$ To visualize horizontal shifts, we can observe the general graph of the parent function $f\left(x\right)={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ and for $\text{\hspace{0.17em}}c>0\text{\hspace{0.17em}}$ alongside the shift left, $\text{\hspace{0.17em}}g\left(x\right)={\mathrm{log}}_{b}\left(x+c\right),$ and the shift right, $\text{\hspace{0.17em}}h\left(x\right)={\mathrm{log}}_{b}\left(x-c\right).$ See [link] .

## Horizontal shifts of the parent function y = log b ( x )

For any constant $\text{\hspace{0.17em}}c,$ the function $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{b}\left(x+c\right)$

• shifts the parent function $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ left $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ units if $\text{\hspace{0.17em}}c>0.$
• shifts the parent function $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ right $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ units if $\text{\hspace{0.17em}}c<0.$
• has the vertical asymptote $\text{\hspace{0.17em}}x=-c.$
• has domain $\text{\hspace{0.17em}}\left(-c,\infty \right).$
• has range $\text{\hspace{0.17em}}\left(-\infty ,\infty \right).$

Given a logarithmic function with the form $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{b}\left(x+c\right),$ graph the translation.

1. Identify the horizontal shift:
1. If $\text{\hspace{0.17em}}c>0,$ shift the graph of $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ left $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ units.
2. If $\text{\hspace{0.17em}}c<0,$ shift the graph of $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ right $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ units.
2. Draw the vertical asymptote $\text{\hspace{0.17em}}x=-c.$
3. Identify three key points from the parent function. Find new coordinates for the shifted functions by subtracting $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ from the $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ coordinate.
4. Label the three points.
5. The Domain is $\text{\hspace{0.17em}}\left(-c,\infty \right),$ the range is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ and the vertical asymptote is $\text{\hspace{0.17em}}x=-c.$

## Graphing a horizontal shift of the parent function y = log b ( x )

Sketch the horizontal shift $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{3}\left(x-2\right)\text{\hspace{0.17em}}$ alongside its parent function. Include the key points and asymptotes on the graph. State the domain, range, and asymptote.

Since the function is $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{3}\left(x-2\right),$ we notice $\text{\hspace{0.17em}}x+\left(-2\right)=x–2.$

Thus $\text{\hspace{0.17em}}c=-2,$ so $\text{\hspace{0.17em}}c<0.\text{\hspace{0.17em}}$ This means we will shift the function $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{3}\left(x\right)\text{\hspace{0.17em}}$ right 2 units.

The vertical asymptote is $\text{\hspace{0.17em}}x=-\left(-2\right)\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}x=2.$

Consider the three key points from the parent function, $\text{\hspace{0.17em}}\left(\frac{1}{3},-1\right),$ $\left(1,0\right),$ and $\text{\hspace{0.17em}}\left(3,1\right).$

The new coordinates are found by adding 2 to the $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ coordinates.

Label the points $\text{\hspace{0.17em}}\left(\frac{7}{3},-1\right),$ $\left(3,0\right),$ and $\text{\hspace{0.17em}}\left(5,1\right).$

The domain is $\text{\hspace{0.17em}}\left(2,\infty \right),$ the range is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ and the vertical asymptote is $\text{\hspace{0.17em}}x=2.$

exercise 1.2 solution b....isnt it lacking
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Almighty formula or by factorization...or by graphical analysis
Damian
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yes am hia
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sebd me some questions about anything ill solve for yall
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x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
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x=7
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3x-4x-7=0 -x=7 x=-7
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