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The definition and properties of logarithmic functions with base a, where 0<a and a≠1. It covers topics such as changing exponential statements to logarithmic statements and vice versa, evaluating logarithmic expressions, determining the domain of a logarithmic function, and graphing logarithmic functions. examples and equations to illustrate the concepts. suitable for students studying precalculus or math analysis.
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The logarithmic function with base a, where a 0 and a 1 ,is denoted by y =logax (read as “y is the logarithm with
base a of x”) and is defined by y =logaxif and only if x =ay. The domain of the function y =logax is x |x 0 .
As this definition illustrates, a logarithm is a name for a certain exponent. So loga x represents the exponent to which a must be raised to obtain x. So, when you need to evaluate loga x,think to yourself “a raised to what power gives me x?”
You should be able to change from exponential to logarithmic form and from logarithmic to exponential form.
Example 1: Change each exponential statement to Change each logarithmic statement to an equivalent logarithmic statement. an equivalent exponential statement. a) If a^5 = 32 , then loga 32 = 5. a) If log 8 5 =x,then 8 x^ = 5. b) If ( 6. 4 )^2 =x,then log 6. 4 x= 2. b) If logx 7 = 4 ,thenx^4 = 7.
To find the exact value of a logarithm, write the logarithm in exponential notation using the fact that y =logaxis
equivalent to ay =x,and use the fact that if a x^ =ay,thenx =y.
Example 2 : Find the exact value oflog 381. To evaluate log 81 3 ,think “3 raised to what power yields 81?” Then, let y =log 381. The exponential form is 3 y^ = 81. 3 y^ = 34 y = 4 Thus,log 3 81 = 4.
The logarithmic function y =logax has been defined as the inverse of the exponential functiony =ax.
Thus, if f (x) =ax,then f −^1 (x)=logax.
Based on the discussion in Section 5.2, you know that for a function f and its inversef −^1 ,
Domain of f = Range of f −^1 and Range of f = Domain of f −^1.
So, it follows that the Domain of the logarithmic function = Range of the exponential function =( 0 ,) and the Range of the logarithmic function = Domain of the exponential function =( −,).
Summarizing some properties of the logarithmic function: y =logax (defining equation: x = ay) Domain: 0 x Range: − y
The domain of a logarithmic function consists of the positive real numbers, so the argument of a logarithmic function must be greater than zero.
Example 3: Find the domain of each logarithmic function. a) f (x)=log 4 ( 2 −x) Need 2 −x 0 x 2 Domain =^ x |x 2 or ( −, 2 )
–4 –3 –2 –1 1 2 3 4 x
1
2
3
4
y
b) g( x) log (x^24 ) 3
Needx^2 − 4 0 −x − 2 − 2 x 2 2 x ( x+ 2 )(x− 2 ) 0 x + 2 − + + Set ( x+ 2 )(x− 2 )= 0 x − 2 − − + x + 2 = 0 or x − 2 = 0 x =− 2 or x = 2 Domain = x |x− 2 or x 2 or ( − −, 2) ( 2, )
c) 1 4
h( x ) =log x
Since x 0,provided x 0,the domain of h( x )consists of all real numbers except zero, or using interval notation, ( − , 0) (0, ).
Since exponential functions and logarithmic functions are inverses of each other, the graph of the logarithmic function y =logax is the reflection about the line y =xof the graph of the exponential function y =a ,x as shown below.
If base a is such that 0 a 1 : Example 4: Graph f (x)=( 0. 5 )x and f −^1 (^ x)=log 0. 5 x.
x f (x)=( 0. 5 )x x f −^1 (^ x)=log 0. 5 x − 2 4 0.25 2 − 1 2 0.5 1 0 1 1 0 1 0.5 2 − 1 2 0.25 4 − 2 0.64118574 0.64118574 0.64118574 0.
( x+ 2 )(x− 2 ) + +
Want x^2 − 4 0
f(x)
f −^1 (x)
y =x
–3 3 6 9 x
1
2
3
4
5
g ( x )
–9 –6 –3 3 x
1
2
3
4
5
f ( x )
–3 3 6 9 x
1
2
3
4
5
g ( x )
–3 3 6 9 x
1
2
3
4
5
f ( x )
Domain of f(x):( −, 0 ) Range of f(x):( −,) VA:x = 0
Domain: x − 4 0 x 4
Range:( −,) VA:x = 4
x = 0
x = 4
x = 0
x = 0
x = 0
the number 10 ( log 10 ), the result is the common logarithm function. If the base a of the logarithmic function is not
indicated, it is understood to be 10. That is, y =logxif and only if x =10 .y
Because (^) y =logxand the exponential function (^) y = 10 xare inverse functions, the graph of (^) y =logxcan be obtained
by reflecting the graph of (^) y = 10 xabout the liney =x.
All material has been taken from Precalculus, by M. Sullivan, 10 th^ Edition