Transcript No Slide Title

By Dr. Julia Arnold
Concept 1
The 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
0
1
2
-1
-2
2x
1
2
4
1/2
1/4
This is the graph of 2x
y
x
Pi or  is what is called a transcendental number
( which means it is not the root of some number).

anyroot
any#
e 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.
ex is called the natural exponential function.
Concept 2
The 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 3
If the exponential is 10x...
its inverse is log x
Read log x
Base 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 x
Read 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 3
How to graph the logarithmic
function.
Let’s look at the two graphs of the exponential and the
logarithmic function:
Goes through (0,1) which means
1 = 20
y
x
The range is all positive real
numbers.
f(x)=2x
Goes through (1,0) which means
0 = log2(1)
y
x
f(x)=log2(x)
The domain is all real numbers.
The domain 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
y
x
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
0
1
2
-1
-2
2x
y
1
2
4
1/2
1/4
x
f(x)=2x
Switch
x
1
2
4
1/2
1/4
log2(x)
0
1
2
-1
-2
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.
y
x
This is the
easy way to do
a log graph.
Concept 4
Changing from exponential
form to logarithmic form.
First task is to be able to go from exponential form to
logarithmic form.
x = ay becomes y = loga(x)
Both of these are referred to as bases
Y is the exponent
on the left.
Thus, 25 = 32 becomes
Logs are = to the exponent
on the right.
log2(32)=5
Read: log 32 base 2 =‘s 5
Thus, 25 = 32 becomes
log2(32) = 5
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
103 = 1000 becomes log (1000) = 3
e1 = e becomes
ln (e) = 1
Concept 5
Four log properties.
There are a few truths about logs which we will call
properties:
1. loga(1) = 0 for any a > 0 and not equal to 1
because a0=1 (exponential form of log form)
2. loga(a) = 1 for any a > 0 and not equal to 1
because a1=a (exponential form of log form)
3. loga(ax) = x for any a > 0 and not equal to 1
because 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 = y
you 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)
-2
-3
Click on the green arrow of the correct
answer above.
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 6
The three expansion
properties of logs.
The 3 expansion properties of logs
1. loga(uv) = logau + logav
Proof: Set logau =x and logav =y then change to
exponential form.
ax = u and ay = v.
ax+y =ax ay = uv so, write
ax+y = uv in log form
loga(uv) = x + y
but that’s logau =x and logav =y , so write
loga(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
u
2. log a  log a u  log a v
v
3.
loga un  n loga u
Example 1
Expand to single log expressions:
log10z
Applying property 1
log 10z  log 10  log z  1  log z
Log 10 = 1 from the 2nd property which we
had earlier.
Example 2
Expand to single log expressions:
xy
ln
e
Applying property 2
xy
ln
 ln( xy )  ln e
e
Applying property 1 and from before ln e = 1
xy
ln
 ln( xy )  ln e  ln x  ln y  1
e
Example 3
Expand to single log expressions:
2
x
ln
3
y
First change the radical to an exponent.
x 
x
ln 3  ln 3 
y
y 
2
2
1
2
Next, apply property 3
1
2
x 
x
1 x 
ln 3  ln 3   ln 3 
y
2 y 
y 
2
2
2
1
2
x 
x
1 x 
ln 3  ln 3   ln 3 
y
2 y 
y 
2
2
2
Next, apply property 2 for quotients
1
2
x 
x
1 x 
ln 3  ln 3   ln 3  
y
2 y 
y 
2
2
1
2
3
ln
x

ln
y


2
2
1
2
2
2



x
x
1
x 
ln 3  ln 3   ln 3  
y
2 y 
y 
2
1
2
3
ln x  ln y 

2
Next, apply property 3to the exponentials.
1
2
2
2



x
x
1
x 
ln 3  ln 3   ln 3  
y
2 y 
y 
2
1
1
2
3
ln x  ln y   2 ln x  3 ln y 

2
2
Now they
are single
logs.
Your turn:
Expand to single logs:
2
log6 6 z
3
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 which
will expand to:
3
log6 6 z  log6 6  log6 z
2
2
Now we use property 3
This is the final answer.
3
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 expand
further to:
3
3
log6 6 z  log6 6  log6 z 
2 log6 6  3 log6 z
2
2
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 log6 6  3 log6 z 
2(1)  3 log6 z  2  3 log6 z
log6 6  1
loga a  x
x
is the property.
Or from the beginning of the problem you
could have said:
log6 6  2
2
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.
u
log a u  log a v  log a
v
3.
n loga u  loga un
Condense to a single log:
2 log x  log y
Always begin by reversing property 3
n loga u  loga u
n
logx  log y
2
Next use property 2
x
log
y
2
u
log a u  log a v  log a
v
which is a single log
Condense to a single log:
2 lnx  1  3 lnx  1
reversing property 3
lnx  1  lnx  1
2
3
Next use property 2

x  1
ln
3
x  1
2
which is a single log
Concept 7
Finding logs on your calculator for
any 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 pressing
the 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
Change to exponential form 2x = 5
Take the log (base 10 ) of both sides.
log 2x = log 5
x log 2 = log 5 using the 3rd expansion property
thus x = log 5
log 2
This shows us how we can create the change of base
formula:
logc b
loga b 
logc a
Changes the base to c.
The change of base formula:
logc b
loga b 
logc a
We are given base a, and we
change to base c.
Example 1
Find the following value using
a calculator:
log6 8
Since the calculator is built to
find base 10 or base e, choose
either one and use the change
of base formula.
log 8 ln 8
log 6 8 
or
 1160558422
.
log 6 ln 6
Some problems can be done without a
calculator, but not all.
1. Find log3 81
log3 81  log3 34  4 using the property
loga a  x
x
1
2. Find log 1
2 4
2
1
1
log 1  log 1    2
2 4
2 2
using the same
property
Some problems can be done without a
calculator, but not all.
1. Find log2 32
log2 32  log2 2  5
5
log2 32  log2 25  2
Click on the correct slide to advance to the next slide.
Click on the wrong side and you will remain here.
Right!
log2 32  log2 25  5
Find ln 10
ln 10  1
ln 10  2.30
Right!
log10  1
while
ln 10  2.30
as found using a calculator.
Which of the following is false
concerning log3 36
log3 36  2log3 3  log3 2
log3 36  2  log3 4
log36
log3 36 
log3
None are false
Because of the log properties all of the
statements below are true
log3 36  log3 3222  log3 32  log3 22  2log3 3  log3 2
 ln 2 
2log3 3  log3 2  21 
  3.26
 ln 3 
2
2
log3 36  log3 3  log3 2  2  log3 4
log 4
2  log3 4  2 
 3.26
log3
log36 Uses the change of base
log3 36 
log3
formula
log36
 3.26
log3
Concept 8
Solving exponential equations with
logs
e  15
x
Solve
Take the ln of both sides.
ln e  ln 15
x  ln 15
x
Solve
Property: lnex = x
6(10 )  42
x
Isolate the exponential expression.
10  7 Take the log base 10 of both sides.
log10x  log7
x
x  log7
Solve
15e  6  1
x
Isolate the exponential expression.
15e x  5
1
x
Take the ln of both sides
e 
3
1
ln e  ln  Property: lnex = x
3
x
1
x  ln   ln 1  ln 3  0  ln 3   ln 3
3
Solve
6(10x )  42
Isolate the exponential expression.
10x  7
Take the log base 10 of both sides.
log10x  log7
x  log7
Concept 9
Solving logarithm equations with
exponents
Solve
ln(x  1)  ln 3
Remember property 4. If loga x = loga y , then x = y.
That property applies in this problem, thus
x-1=3
x=4
Solve ln 4x  1
Change to exponential form
e 1  4x
e  4x
e
x
4
Solve ln(x  1)2  2
Apply the expansion property for exponents.
2 ln(x  1)  2
Isolate the ln expression.
2
ln( x  1)   1
2
Change to the exponential form
(x  1)  e1
Solve for x
x  e 1
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