Logarithmic Functions
By
Dr. Julia Arnold and Ms. Karen Overman
using Tan’s 5th edition Applied Calculus for the
managerial , life, and social sciences text
Concept 1
The logarithmic function is the
inverse of the exponential function.
Recall that an Inverse Function reverses the ordered pairs
which belong to the original function. 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)
Note:Base 10 is considered the common base and thus log x is
the common log and as such the base is omitted and understood
to be 10.
If the exponential is y=2x, then its inverse is y=log2 (x).
(read y = log x base 2 )
Are you getting the hang of it??
If the exponential is y=3x ...
its inverse is y=log3(x)
If the exponential is y=10x...
its inverse is y=log x
Note: 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
ex...
its inverse is ln x
(read l n x or natural log x)
Note: 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 2
How to graph the logarithmic
function.
Remember this graph from the last section on
exponential functions?
We used a table of values to graph the exponential y = 2x.
x
2x
0
1
2
-1
-2
1
2
4
1/2
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.
y = log2(x) is the INVERSE of the function y = 2x.
To create the graph we can just switch the ordered pairs
in the left table and create the log graph for base 2.
x
2x
x
log2(x)
0
1
2
-1
-2
1
2
4
1/2
1/4
1
2
4
1/2
1/4
0
1 This is the easy
way to do a log
2
graph.
-1
-2
Switch
y
y
x
y=2x
x
y = log2(x)
Let’s look at the two graphs:
The exponential function, f(x)=2x
goes through (0, 1), since 1 = 20.
y
x
The domain is all real numbers.
The range is all positive real numbers.
f(x)=2x
The logarithmic function, f(x)=log2(x)
goes through (1, 0) which means
log2(1) = 0.
y
x
The domain is all positive real numbers.
f(x)=log2(x)
The range is all real numbers.
Concept 3
Converting between exponential
form and logarithmic form.
Exponential Form
x = ay
Logarithmic Form
becomes
y = loga(x)
In both forms, a is called the base.
In logarithmic form,
y equals the log or
the log equals the
exponent.
In exponential form,
y is the exponent.
25 = 32
becomes
log2(32)=5
Read: log 32 base 2 =‘s 5
Let’s practice converting from exponential form
to logarithmic form.
We saw that, 25 = 32 becomes
log2(32) = 5
So then, 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
Logarithmic Form
y = loga(x)
Exponential Form
becomes
x = ay
Recall, in both forms, a is called the base. In
logarithmic form, the log equals the exponent, thus y
is the exponent in exponential form.
For example,
log2(32)=5
becomes
25 = 32
Let’s practice converting from logarithmic form
to exponential form.
We saw that, log2(32) = 5 becomes
25 = 32
So then, log3(81) = 4 becomes
34 = 81
log2(1/8) = -3 becomes
2-3 = 1/8
log5(1/25) = -2 becomes
5-2 = 1/25
log2(1) = 0 becomes
20 =
1
103 = 1000
log (1000) = 3 becomes
ln (e) = 1 becomes
e1 = e
Concept 4
Properties for Logs.
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:
If log3x = log3 4, then x = 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 and the rules for exponents:
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 24 = -4
which totals to -3
Go back.
Way to go!
Using properties 1,2 and 3:
ln 1 = 0
log 10 = 1
-log2 24 = -4
which totals to -3
Concept 5
Expansion properties of logs.
Expansion properties of logs
1. loga(uv) = logau + logav
u
2. log a  log a u  log a v
v
3.
log a u n  nlog a u
The first expansion property is proven below.
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?
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 the logarithm completely.
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 the logarithm completely.
2
x
ln 3
y
First change the radical to an exponent.
1
2
2 2
x 
x
ln 3  ln 3 
y
y 
Wait, we’re not done yet! Next, apply property 3.
Example 3 continued:
After applying property 3.
ln
 x2 
x2
3  ln
3
y
 y 
1
2
1  x2 

ln 3 
2  y 
Next, apply property 2 for quotients
ln
 x2 
x2
3  ln
3
y
 y 
1
ln x 2  ln y 3 

2
1
2
1  x2 

ln 3  
2  y 
Example 3 continued:
After applying property 2 for quotients.
Next, apply property 3to the exponentials.
x 
x
ln
3  ln
3
y
y 
2
2
1
2
1 x 
 ln 3  
2 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 the logarithm completely:
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  log 6 z 
2
2
2 log6 6  3 log6 z
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 log 6 6  3 log 6 z 
2(1)  3 log 6 z  2  3 log 6 z
log a a  x
x
is the property.
log 6 6  1
Or from the beginning of the problem you
could have said:
log6 6  2
2
We can use the same three expansion properties
of logs to take an expanded log and condense it
back to a single log expression.
1.
logau + logav =loga(uv)
u
2. log a u  log a v  log a
v
3.
nlog a u  log a u
n
Here’s an example:
Condense this expression into a single log.
2 log x  log y
Always begin by reversing property 3:
n log a u  log a u
n
log x  log y
2
Next use property 2:
x2
log
y
u
log a u  log a v  log a
v
(Now we have a single log.)
Another example:
Condense the expression into a single log.
2 ln x  1  3 ln x  1
First, reverse property 3.
ln x  1  ln x  1
2
3
Next, use property 2.

x  1
ln
3
x  1
2
(Now we have a single log.)
Concept 6
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 using the log key
or the
ln(of a number) to the base e by using the ln key.
What if you want to find the value of a log to a different
base?
For example, how can we find the value of log25 ?
How can we find the value of log25 ?
1. Set log25 = x
2. Change to exponential form 2x = 5
3. Take the log (base 10 ) of both sides.
log 2x = log 5
4. 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:
log c b
log a b 
Changes the base to c.
log c a
The change of base formula:
log c b
log a b 
log c a
We are given base a, and we
change to base c.
Example 1:
Find the following value using a calculator:
log 6 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
log6 8 
or
 1160558422
.
log 6 ln 6
Some problems can be done without a calculator,
but not all.
1. Find
log 3 81
log 3 81  log 3 34  4
2. Find
1
log 1
2 4
2
1
1
log 1  log 1    2
2 4
2 2
Here’s a problem that can be done without a calculator.
1. Find
log 2 32
log 2 32  log 2 2  5
5
log 2 32  log 2 25  2
Click on the correct slide to advance to the next slide.
Click on the wrong side and you will remain here.
Right!
log 2 32  log 2 25  5
Now find ln 10. Do you need a calculator?
ln 10  1
ln 10  2.30
Click on the correct slide to advance to the next slide.
Click on the wrong side and you will remain here.
Right!
ln 10  2.30
as found using a calculator.
Which of the following is false concerning:
log 3 36
log 3 36  2log 3 3  log 3 2
log 3 36  2  log 3 4
log 36
log 3 36 
log 3
None are false
Click on the correct slide to advance to the next slide.
Click on the wrong side and you will remain here.
Because of the log properties all of the
statements were true.
log 3 36  log 3 32 2 2  log 3 32  log 3 2 2  2log 3 3  log 3 2
 ln2 
2log 3 3  log 3 2  21 
  3.26
 ln3 
log 3 36  log 3 32  log 3 22  2  log 3 4
log 4
2  log 3 4  2 
 3.26
log 3
log 36 Uses the change of base formula
log 3 36 
log 3
log 36
 3.26
log 3
Concept 7
Solving exponential equations by
using logs.
Solve
e x  15
Since you cannot get the bases the same as we did in the
previous lesson (5.1), take the ln of both sides.
ln e  ln 15
x  ln 15
x
Use the property: lnex = x
Solve
6(10 x )  42
First, isolate the exponential expression.
10 x  7
Take the log base 10 of both sides.
log 10 x  log 7
x  log 7
Use the property: log 10x = x
Solve
15e x  6  1
Isolate the exponential expression.
15e x  5
1
e 
3
x
Take the ln of both sides.
1
ln e  ln   Using the property: lnex = x
3
x
1
x  ln    ln 1  ln 3  0  ln 3   ln 3
3
Concept 8
Solving logarithmic 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  4x
1
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
Review
We covered a lot of information about logs. Here is a list of the
topics we studied. Hopefully you took some notes along the way!!
1. The logarithmic function is the inverse of the exponential function.
2. Graphing a logarithmic function.
3. Converting between logarithmic form and exponential form.
4. The properties of logarithms.
5. The expansion properties of logarithms and using them.
6. Finding the value of logarithms using your calculator and the change
of base formula.
7. Solving exponential equations by using logarithms. (When we cannot
get the bases the same.)
8. Solving logarithmic equations by isolating the logarithm and
converting to exponential form.