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The exponential function f with base a is denoted by f(x) = a x where a > 0 a 1, and x is any real number. To graph a specific exponential function we will use a table of values. x 2 x /2 -2 1/4 This is the graph of 2 x
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By Dr. Julia Arnold
Concept 1The Exponential Function
The exponential function f with base a is denoted by f(x) = ax where a > 0 a 1, and x is any real number.
To graph a specific exponential function we will use a table of values.
x 2x 0 1 1 2 2 4 -1 1/2 -2 1/4
x
y
This is the graph of 2x
Pi or is what is called a transcendental number ( which means it is not the root of some number).
e
ex is called the natural exponential function.
is another such number. On the graphing calculator, you can find e by pushing the yellow 2nd function button and the Ln key. On the display you will see e^( Type 1 and close parenthesis. Thus e^( 1). Press Enter and you will see 2.718281828 which represents an approximation of e.
anyrootany#
Concept 2The exponential function is the inverse of the logarithm function
Recall that Inverse Functions reverse the ordered pairs which belong to functions. i.e. (x,y) becomes (y,x)
The log function is the inverse of the exponential function.
If the exponential is 2x, then its inverse is log2 (x) (read log x base 2 ).If the exponential is 3x... its inverse is log3(x)
Read log x base 3If the exponential is 10x... its inverse is log x
Read log xBase 10 is considered the common base and thus log x is the common log and as such the base is omitted.
If the exponential is 2x=y, then its inverse is log2 (x)=y (read log y base 2 =‘s x ).
If the exponential is 3x... its inverse is log3(x)
If the exponential is 10x... its inverse is log x
Base 10 is considered the common base and thus log y is the common log and as such the base is omitted.
If the exponential is ex... its inverse is ln xRead l n x
Base e is considered the natural base and thus ln x is the natural log and is written ln to distinguish it from log.
Concept 3How to graph the logarithmicfunction.
Let’s look at the two graphs of the exponential and thelogarithmic function:
x
y
x
y
f(x)=2x
f(x)=log2(x)
Goes through (0,1) which means 1 = 20
Goes through (1,0) which means 0 = log2(1)
The domain is all real numbers.
The domain is all positive real numbers.
The range is all positive real numbers.
The range is all real numbers.
Remember this slide?We used a table of values to graph the exponential y = 2x.
x 2x 0 1 1 2 2 4 -1 1/2 -2 1/4
x
y
f(x)=2x
Since we know that the y = log2(x) is the inverse of the function above, we can just switch the ordered pairs in the table above and create the log graph for base 2.
x 2x 0 1 1 2 2 4 -1 1/2 -2 1/4
x
y
f(x)=2x
y = log2(x) is the inverse of the function y = 2x. To create the graph we can just switch the ordered pairs in the table left and create the log graph for base 2.
x log2(x)1 0 2 14 21/2 -11/4 -2
Switch
x
y
x
y
x
y
x
y
x
y
This is the easy way to do a log graph.
Concept 4Changing from exponentialform to logarithmic form.
First task is to be able to go from exponential form tologarithmic form.
x = ay becomes y = loga(x)
Both of these are referred to as bases
Y is the exponent on the left.
Logs are = to the exponent on the right.
Thus, 25 = 32 becomes log2(32)=5Read: log 32 base 2 =‘s 5
Thus, 25 = 32 becomes log2(32) = 5
103 = 1000 becomes log (1000) = 3
e1 = e becomes ln (e) = 1
34 = 81 becomes log3(81) = 4
2-3 = 1/8 becomes log2(1/8) = -3
5-2 = 1/25 becomes log5(1/25) = -2
20 = 1 becomes log2(1) = 0
Concept 5Four log properties.
There are a few truths about logs which we will callproperties:
1. loga(1) = 0 for any a > 0 and not equal to 1because a0=1 (exponential form of log form)
2. loga(a) = 1 for any a > 0 and not equal to 1because a1=a (exponential form of log form)
3. loga(ax) = x for any a > 0 and not equal to 1because ax=ax (exponential form of log form)
4. If loga x = loga y , then x = y.
Practice Problems
1. Solve for x: log3x = log3 4
X = 34
X = 4
Click on the green arrow of the correct answer above.
No, x = 34 is not correct. Use the 4th property:4. If loga x = loga y , then x = y. log3x = log3 4, then x = 4
Go back.
Way to go! Using the 4th property:4. If loga x = loga y , then x = yyou concluded correctly that x = 4 for log3x = log3 4.
Practice Problems
2. Solve for x: log21/8 = x
X = 3
X = -3
Click on the green arrow of the correct answer above.
No, x = 3 is not correct. Use the 3rd property:3. loga(ax) = x log21/8 = log2 8-1 = log2 (23)-1 =
log2 2-3 then x = -3 since the 2’s make a match.
Go back.
Way to go!
Using the 3rd property:3. loga(ax) = x log21/8 = log2 8-1 = log2 (23)-1 =
log2 2-3 then x = -3 since the 2’s make a match.
Practice Problems
3. Evaluate: ln 1 + log 10 - log2(24)
Click on the green arrow of the correct answer above.
-3
-2
No, -2 is not correct. Using properties 1,2 and 3:ln 1 = 0 log 10 = 1 log2 2-4 = -4 which totals to -3
Go back.
Way to go!
Using properties 1,2 and 3:ln 1 = 0 log 10 = 1 log2 2-4 = -4 which totals to -3
Concept 6The three expansion properties of logs.
The 3 expansion properties of logs
1. loga(uv) = logau + logavProof: Set logau =x and logav =y then change to exponential form. ax = u and ay = v. ax+y =ax ay = uv so, writeax+y = uv in log formloga(uv) = x + ybut that’s logau =x and logav =y , so writeloga(uv) = logau + logav which is the result we were looking for.Do you see how this property relates to the
exponential property?
The 3 expansion properties of logs
1. loga(uv) = logau + logav
2. vuvu
aaa logloglog
3. unu an
a loglog
Example 1 Expand to single log expressions:
lo g 1 0 zApplying property 1
lo g lo g lo g lo g1 0 1 0 1z z z
Log 10 = 1 from the 2nd property which wehad earlier.
Example 2 Expand to single log expressions:
lnxye
Applying property 2
ln ln ( ) lnxye
xy e
Applying property 1 and from before ln e = 1
ln ln ( ) ln ln lnxye
xy e x y 1
Example 3 Expand to single log expressions:
lnxy
2
3First change the radical to an exponent.
ln lnxy
xy
2
3
2
3
12
Next, apply property 3
ln ln lnxy
xy
xy
2
3
2
3
12 2
3
12
ln ln lnxy
xy
xy
2
3
2
3
12 2
3
12
Next, apply property 2 for quotients
ln ln ln
ln ln
xy
xy
xy
x y
2
3
2
3
12 2
3
2 3
12
12
ln ln ln
ln ln
xy
xy
xy
x y
2
3
2
3
12 2
3
2 3
12
12
Next, apply property 3to the exponentials.
ln ln ln
ln ln ln ln
xy
xy
xy
x y x y
2
3
2
3
12 2
3
2 3
12
12
12
2 3
Now theyare singlelogs.
Your turn:Expand to single logs:
lo g 62 36 z
The first step is to use property 1
The first step is to use property 3
No, incorrect, return to the previous slide.
The first step is to use property 1 whichwill expand to:
lo g lo g lo g62 3
62
636 6z z
Now we use property 3
This is the final answer.
No, this is not the final answer.Return to the previous slide and click on the correct answer.
Yes, we now use property 3 to expandfurther to:lo g lo g lo g
lo g lo g6
2 36
26
3
6 6
6 62 6 3
z zz
This is the final answer.This is still not the final answer.
Nope, we are not done yet.Return
Whenever the base of the log matches the number you are taking the log of, the answer is the exponent on the number which is 1 in this case.
2 6 32 1 3 2 3
6 6
6 6
lo g lo g( ) log lo g
zz z
lo g
lo g6 6 1
axa x is the property.
lo g 626 2
Or from the beginning of the problem youcould have said:
We can use the same 3 expansion properties of logs to take an expanded log and condense it back to a single log expression.
1. logau + logav =loga(uv)
2. vuvu aaa logloglog
3. naa uun loglog
Condense to a single log:
yx2 loglog Always begin by reversing property 3
naa uun loglog
yx2 loglog
Next use property 2 vuvu aaa logloglog
yx2
log which is a single log
Condense to a single log: 1x31x2 lnln
reversing property 3
Next use property 2
which is a single log
32 1x1x lnln
3
2
1x1x
ln
Concept 7Finding logs on your calculator forany base number.
On your graphing calculator or scientific calculator, you may find the value of the log (of a number) to the base 10 or the ln(of a number) to the base e by simply pressingthe appropriate button.What if you want to find the value of a log to a different base?
How can we find for example, the log25
How can we find for example, the log25
Set log25 = x
25
loglog
This shows us how we can create the change of baseformula:
abb
c
ca log
loglog Changes the base to c.
Change to exponential form 2x = 5Take the log (base 10 ) of both sides.log 2x = log 5
x log 2 = log 5 using the 3rd expansion propertythus x =
abb
c
ca log
loglog
The change of base formula:
We are given base a, and we change to base c.
Example 1Find the following value usinga calculator:
lo g 6 8
Since the calculator is built to find base 10 or base e, choose either one and use the change of base formula.
lo glo glo g
lnln
.6 886
86
1 1 6 0 5 5 8 4 2 2 or
Some problems can be done without acalculator, but not all.
813log1. Find4381 4
33 logloglo g a
xa x using the property
2. Find
using the same property
41
21
log
221
41 2
21
21
loglog
Some problems can be done without acalculator, but not all.
322log1. Find5232 5
22 loglog
2232 522 loglog
Click on the correct slide to advance to the next slide. Click on the wrong side and you will remain here.
5232 522 loglog
Right!
Find ln 10
30210 .ln
110ln
110log
Right!
while 30210 .ln
as found using a calculator.Which of the following is falseconcerning 363log
23236 333 logloglog
4236 33 loglog
336363 log
loglog
None are false
232232336 332
32
322
33 loglogloglogloglog
Because of the log properties all of the statements below are true
2633212232 33 .
lnlnloglog
422336 32
32
33 loglogloglog
26334242 3 .
logloglog
336363 log
loglog Uses the change of base formula
263336 .
loglog
Concept 8Solving exponential equations with logs
Solve 15ex Take the ln of both sides.
15ex lnln 15x ln
Property: lnex = x
Solve 42106 x )(Isolate the exponential expression.
Take the log base 10 of both sides.710x
7x710x
logloglog
Solve 16e15 x Isolate the exponential expression.
Property: lnex = x31e
5e15x
x
Take the ln of both sides
31ex lnln
3303131x lnlnlnlnln
Solve 42106 x )(Isolate the exponential expression.
Take the log base 10 of both sides.710x
7x710x
logloglog
Concept 9Solving logarithm equations with exponents
Solve 31x ln)ln( Remember property 4. If loga x = loga y , then x = y.That property applies in this problem, thusx - 1 = 3x = 4 Solve 1x4 ln
Change to exponential form
x4e
x4ex4e1
Solve 21x 2 )ln(
Apply the expansion property for exponents.21x2 )ln(
1221x )ln(
Isolate the ln expression.
1e1x )(
Change to the exponential form
1ex Solve for x
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