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Section exercises

Verbal

The inverse of every logarithmic function is an exponential function and vice-versa. What does this tell us about the relationship between the coordinates of the points on the graphs of each?

Since the functions are inverses, their graphs are mirror images about the line y = x . So for every point ( a , b ) on the graph of a logarithmic function, there is a corresponding point ( b , a ) on the graph of its inverse exponential function.

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What type(s) of translation(s), if any, affect the range of a logarithmic function?

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What type(s) of translation(s), if any, affect the domain of a logarithmic function?

Shifting the function right or left and reflecting the function about the y-axis will affect its domain.

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Consider the general logarithmic function f ( x ) = log b ( x ) . Why can’t x be zero?

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Does the graph of a general logarithmic function have a horizontal asymptote? Explain.

No. A horizontal asymptote would suggest a limit on the range, and the range of any logarithmic function in general form is all real numbers.

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Algebraic

For the following exercises, state the domain and range of the function.

f ( x ) = log 3 ( x + 4 )

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h ( x ) = ln ( 1 2 x )

Domain: ( , 1 2 ) ; Range: ( , )

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g ( x ) = log 5 ( 2 x + 9 ) 2

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h ( x ) = ln ( 4 x + 17 ) 5

Domain: ( 17 4 , ) ; Range: ( , )

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f ( x ) = log 2 ( 12 3 x ) 3

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For the following exercises, state the domain and the vertical asymptote of the function.

f ( x ) = log b ( x 5 )

Domain: ( 5 , ) ; Vertical asymptote: x = 5

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g ( x ) = ln ( 3 x )

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f ( x ) = log ( 3 x + 1 )

Domain: ( 1 3 , ) ; Vertical asymptote: x = 1 3

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f ( x ) = 3 log ( x ) + 2

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g ( x ) = ln ( 3 x + 9 ) 7

Domain: ( 3 , ) ; Vertical asymptote: x = 3

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For the following exercises, state the domain, vertical asymptote, and end behavior of the function.

f ( x ) = ln ( 2 x )

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f ( x ) = log ( x 3 7 )

Domain: ( 3 7 , ) ;
Vertical asymptote: x = 3 7 ; End behavior: as x ( 3 7 ) + , f ( x ) and as x , f ( x )

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h ( x ) = log ( 3 x 4 ) + 3

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g ( x ) = ln ( 2 x + 6 ) 5

Domain: ( 3 , ) ; Vertical asymptote: x = 3 ;
End behavior: as x 3 + , f ( x ) and as x , f ( x )

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f ( x ) = log 3 ( 15 5 x ) + 6

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For the following exercises, state the domain, range, and x - and y -intercepts, if they exist. If they do not exist, write DNE.

h ( x ) = log 4 ( x 1 ) + 1

Domain: ( 1 , ) ; Range: ( , ) ; Vertical asymptote: x = 1 ; x -intercept: ( 5 4 , 0 ) ; y -intercept: DNE

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f ( x ) = log ( 5 x + 10 ) + 3

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g ( x ) = ln ( x ) 2

Domain: ( , 0 ) ; Range: ( , ) ; Vertical asymptote: x = 0 ; x -intercept: ( e 2 , 0 ) ; y -intercept: DNE

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f ( x ) = log 2 ( x + 2 ) 5

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h ( x ) = 3 ln ( x ) 9

Domain: ( 0 , ) ; Range: ( , ) ; Vertical asymptote: x = 0 ; x -intercept: ( e 3 , 0 ) ; y -intercept: DNE

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Graphical

For the following exercises, match each function in [link] with the letter corresponding to its graph.

Graph of five logarithmic functions.

For the following exercises, match each function in [link] with the letter corresponding to its graph.

Graph of three logarithmic functions.

f ( x ) = log 1 3 ( x )

B

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h ( x ) = log 3 4 ( x )

C

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For the following exercises, sketch the graphs of each pair of functions on the same axis.

f ( x ) = log ( x ) and g ( x ) = 10 x

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f ( x ) = log ( x ) and g ( x ) = log 1 2 ( x )

Graph of two functions, g(x) = log_(1/2)(x) in orange and f(x)=log(x) in blue.
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f ( x ) = log 4 ( x ) and g ( x ) = ln ( x )

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f ( x ) = e x and g ( x ) = ln ( x )

Graph of two functions, g(x) = ln(1/2)(x) in orange and f(x)=e^(x) in blue.
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For the following exercises, match each function in [link] with the letter corresponding to its graph.

Graph of three logarithmic functions.

f ( x ) = log 4 ( x + 2 )

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g ( x ) = log 4 ( x + 2 )

C

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h ( x ) = log 4 ( x + 2 )

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For the following exercises, sketch the graph of the indicated function.

f ( x ) = log 2 ( x + 2 )

Graph of f(x)=log_2(x+2).
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f ( x ) = 2 log ( x )

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f ( x ) = ln ( x )

Graph of f(x)=ln(-x).
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g ( x ) = log ( 4 x + 16 ) + 4

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g ( x ) = log ( 6 3 x ) + 1

Graph of g(x)=log(6-3x)+1.
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h ( x ) = 1 2 ln ( x + 1 ) 3

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For the following exercises, write a logarithmic equation corresponding to the graph shown.

Use y = log 2 ( x ) as the parent function.

The graph y=log_2(x) has been reflected over the y-axis and shifted to the right by 1.

f ( x ) = log 2 ( ( x 1 ) )

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Use f ( x ) = log 3 ( x ) as the parent function.

The graph y=log_3(x) has been reflected over the x-axis, vertically stretched by 3, and shifted to the left by 4.
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Use f ( x ) = log 4 ( x ) as the parent function.

The graph y=log_4(x) has been vertically stretched by 3, and shifted to the left by 2.

f ( x ) = 3 log 4 ( x + 2 )

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Use f ( x ) = log 5 ( x ) as the parent function.

The graph y=log_3(x) has been reflected over the x-axis and y-axis, vertically stretched by 2, and shifted to the right by 5.
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Technology

For the following exercises, use a graphing calculator to find approximate solutions to each equation.

log ( x 1 ) + 2 = ln ( x 1 ) + 2

x = 2

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log ( 2 x 3 ) + 2 = log ( 2 x 3 ) + 5

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ln ( x 2 ) = ln ( x + 1 )

x 2 .303

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2 ln ( 5 x + 1 ) = 1 2 ln ( 5 x ) + 1

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1 3 log ( 1 x ) = log ( x + 1 ) + 1 3

x 0.472

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Extensions

Let b be any positive real number such that b 1. What must log b 1 be equal to? Verify the result.

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Explore and discuss the graphs of f ( x ) = log 1 2 ( x ) and g ( x ) = log 2 ( x ) . Make a conjecture based on the result.

The graphs of f ( x ) = log 1 2 ( x ) and g ( x ) = log 2 ( x ) appear to be the same; Conjecture: for any positive base b 1 , log b ( x ) = log 1 b ( x ) .

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Prove the conjecture made in the previous exercise.

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What is the domain of the function f ( x ) = ln ( x + 2 x 4 ) ? Discuss the result.

Recall that the argument of a logarithmic function must be positive, so we determine where x + 2 x 4 > 0 . From the graph of the function f ( x ) = x + 2 x 4 , note that the graph lies above the x -axis on the interval ( , 2 ) and again to the right of the vertical asymptote, that is ( 4 , ) . Therefore, the domain is ( , 2 ) ( 4 , ) .

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Use properties of exponents to find the x -intercepts of the function f ( x ) = log ( x 2 + 4 x + 4 ) algebraically. Show the steps for solving, and then verify the result by graphing the function.

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Questions & Answers

preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
A hedge is contrusted to be in the shape of hyperbola near a fountain at the center of yard.the hedge will follow the asymptotes y=x and y=-x and closest distance near the distance to the centre fountain at 5 yards find the eqution of the hyperbola
ayesha Reply
A doctor prescribes 125 milligrams of a therapeutic drug that decays by about 30% each hour. To the nearest hour, what is the half-life of the drug?
Sandra Reply
Find the domain of the function in interval or inequality notation f(x)=4-9x+3x^2
prince Reply
hello
Jessica Reply
Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the high temperature of ?105°F??105°F? occurs at 5PM and the average temperature for the day is ?85°F.??85°F.? Find the temperature, to the nearest degree, at 9AM.
Karlee Reply
if you have the amplitude and the period and the phase shift ho would you know where to start and where to end?
Jean Reply
rotation by 80 of (x^2/9)-(y^2/16)=1
Garrett Reply
thanks the domain is good but a i would like to get some other examples of how to find the range of a function
bashiir Reply
what is the standard form if the focus is at (0,2) ?
Lorejean Reply
a²=4
Roy Reply
hil
Roy Reply
hi
Roy Reply
A bridge is to be built in the shape of a semi-elliptical arch and is to have a span of 120 feet. The height of the arch at a distance of 40 feet from the center is to be 8 feet. Find the height of the arch at its center
Abdulfatah Reply

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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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