Transcript Solving Exponential and Logarithmic Equations

5.6
Solving Exponential
and Logarithmic Equations
Objectives:
Solve exponential and logarithmic equations.
2. Solve a variety of application problems by using
exponential and logarithmic equations.
1.
Methods of Solving Exponential Equations
1.
Rewrite the bases of the powers on both sides so they are the same.
This method is the fastest to use but only works for certain situations. It uses
a fact that we often take for granted. For instance, in the equation 2x = 23, we
would say that x = 3.
If the bases are equal, then the exponents must be equal as well.
If the equation were modified slightly, say 2x – 1 = 23, we would still set the
exponents equal, but now we would say that x – 1 = 3, therefore x = 4.
2.
Take a logarithm of both sides.
This methods always works, and for simple situations it is rather quick, but it
can also be very cumbersome. Especially if variables are on both sides of the
equation.
Example #1
Powers of the Same Base
 Solve the equation and confirm your solution with a graph.
A.
9 3
x
3 
2 x
x 1
3
x 1
x 1
3 3
2x  x  1
2x
x  1
Rewrite both bases to base 3.
Example #1
Powers of the Same Base
 Solve the equation and confirm your solution with a graph.
B.
2
x 3
2 x 3
2
x 3
2
x 3
 
 2 

 2 
 4
1
 
 4
1 x  5
x 5

1
2


2 x 5
2 x  3  2 2 x 10
x 5
Rewrite to base 2, remember fractions have
negative exponents.
x  3  2 x  10
3x  3  10
3x  13
13
x
3
Example #1 continued…
Powers of the Same Base
 Solve the equation and confirm your solution with a graph.
C.
5 25  625
2x
5
2x
x
5 
2 x
 54
52 x 52 x  54
52 x  2 x  54
54 x  54
4x  4
x 1
Rewrite to base 5.
Example #1 continued…
Powers of the Same Base
 Solve the equation and confirm your solution with a graph.
D.
2
2
2
x 2 6
x
 32
 
5 x
x2 6
 2
x2 6
 25x
x 2  6  5 x
x  5x  6  0
2
x  6x  1  0
x  6, x  1
This time the graphs with the intersection
method were too difficult to read, so the
intercept method was used to verify the two
solutions.
Example #2
Logarithms on Both Sides
 Solve the equation. Solutions can be confirmed with a graph.
6 3
x
Problems like this can be confusing because students will recognize that 2(3) = 6.
Unfortunately, that will not work because we have to be able to rewrite both sides to
the same base of a power for the first method to be used. Our only choice then is
the second method.
log 6 x  log 3
x  log 6  log 3
log 3
x
 0.6131
log 6
ln 6 x  ln 3
x  ln 6  ln 3
ln 3
x
 0.6131
ln 6
Notice how common logs and natural logs both work the same way. Choosing which one to
use basically comes down to personal preference.
Example #3
 Solve the equation.
4
2 x 1
This type can be very tricky:
1 x
5
ln 4 2 x 1  ln 51 x
2 x  1ln 4  1  x ln 5
2 x ln 4  ln 4  ln 5  x ln 5
2 x ln 4  x ln 5  ln 5  ln 4
x2 ln 4  ln 5  ln 5  ln 4
ln 5  ln 4
x
 0.6836
2 ln 4  ln 5
1. First take a logarithm of both sides
(common or natural).
2. Swing the exponents in front using
the properties of logarithms.
3. Distribute.
4. Rewrite the expression so that x
terms are on one side of the
equation and everything else is on
the other side. Remember if you
change the side, change the sign.
5. Factor out the GCF of x.
6. Divide.
Example #3
 Solve the equation.
4
2 x 1
1 x
5
ln 5  ln 4
x
 0.6836
2 ln 4  ln 5
Typing the answer in the calculator
can be difficult:
Alternatively you can first simplify the
answer further by using the properties of
logarithms.
ln 5  ln 4
ln 5  4
x

2 ln 4  ln 5 ln 4 2  ln 5
ln 20
ln 20


ln 16  ln 5 ln 16  5
ln 20

 0.6836
ln 80
Example #4
 Solve the equation.
x
e  5e  6
e e  5e   e  6
x
x
x
x
x
e x  e x  5e x  e  x  6e x
1. Multiply both sides by ex.
This will remove the
negative exponent on the e.
e x  x  5e x   x  6e x
e 2 x  5e0  6e x
e 2 x  5(1)  6e x
e 2 x  6e x  5  0
u 2  6u  5  0
2. Rewrite the expression
equal to 0.
3. Substitute u = ex.
Example #4
 Solve the equation.
e  5e
x
x
 6
u  6u  5  0
2
u
62  41 5
21
6
 6  56  6  2 14
u

2
2
u  3  14
e x  3  14

ln e x  ln  3  14



x  ln  3  14  0.2989
1. Use the quadratic formula to
solve for u.
2. Substitute ex back in for u.
3. Take the natural logarithm of
both sides to eliminate the e
on the left side.
4. Only the positive solution
from the quadratic formula is
used because you can’t take
the log of a negative number.
Solving Logarithmic Equations
 To solve a logarithmic equation, it is often necessary to
condense the expression down to a single logarithm on
each side first. This means all the properties of
logarithms will come into play. Additionally, constant
terms and logarithms cannot be on the same side.
 After an individual logarithm is created on each side of
the equation, the logarithm can be removed using its
inverse, a power of the same base.
 Finally, make sure to check for extraneous solutions
since the domain of logarithms are limited to numbers
greater than 0.
Example #5
 Solve the equation and confirm your solution with a graph.
ln(4 x  1)  ln(x  2)  2(ln x)
ln 4 x  1 x  2   ln  x 
2


ln 4 x  7 x  2  ln x
2
ln4 x 2  7 x  2 
ln x 2
e
e
4x  7x  2  x
2
3x  7 x  2  0
2
2
2
First note that 2(ln x) is the same thing
as 2ln x, which had its coefficient
placed up top as an exponent.
Secondly, e was chosen as the base of
the power because it is the inverse of
the natural log. After the logarithms
are removed and the expression is set
equal to 0, we get a quadratic that
cannot factor and will require the
Quadratic Formula.
Example #5
 Solve the equation and confirm your solution with a graph.
ln(4 x  1)  ln(x  2)  2(ln x)
3x 2  7 x  2  0
x
7
7 2  43 2
23
 7  73
x
6
When solving this problem we get two
solutions. It is important to check both
solutions because any value for x (positive
or negative) that makes the resulting
logarithmic expression negative or 0 will
result in no solution.
Take for instance the expression:
ln(x – 5)
If the solutions to an equation with that
expression were 4 and 5, even though
both answers are positive, when plugged
back in we get:
ln(4 – 5) = ln(– 1)
ln(5 – 5) = ln(0)
Both of these expressions are undefined.
Example #5
 Solve the equation and confirm your solution with a graph.
ln(4 x  1)  ln(x  2)  2(ln x)
 7  73
x
6
x  0.25733 or x  2.59067
ln 40.25733   1  ln .02932 
ln 0.25733  2  ln( 2.25733)
ln 4 2.59067   1  ln  11.36268 
ln  2.59067  2  ln  0.59067 
Therefore, the only valid solution is the
positive solution:
 7  73
x
6
x  0.25733
Do not assume this is always the
case. A more efficient way of
checking will be with the graphing
calculator.
Example #5
 Solve the equation and confirm your solution with a graph.
ln(4 x  1)  ln(x  2)  2(ln x)
 7  73
x
6
x  0.25733
A better view:
Note: You will have to move the cursor to select
each curve when using the intersect method.
Secondly, the intersection isn’t showing on this
screen, but it is there.
Example #6
 Solve the equation.
ln(2 x  1)  8  ln(2 x  1)
ln 2 x  1  ln 2 x  1  8
2 ln 2 x  1  8
ln 2 x  1  4
e ln2 x 1  e 4
2x  1  e
4
2x  e4  1
e4  1
x
2
x  26.7991
The trick to this problem is recognizing to
the change the addition of two of the same
expressions into multiplication by 2.
Then both sides are divided by 2 and the
problem is solved fairly straightforward.
Example #6
 Solve the equation.
ln(2 x  1)  8  ln(2 x  1)
ln 2 x  1  ln 2 x  1  8
2 ln 2 x  1  8
Here a second method of solving the
same problem is shown but this time two
possible answers arrive. After checking
for extraneous solutions the same
solution is identified.
ln 2 x  12  8
e
ln2 x 12
 e8
2 x  12  e8
2 x  12
 e8
2 x  1  e 4
 e4  1
x
2
x  26.7991
Example #7
 Solve the equation.
log(x  2)  1  log(x  1)
log  x  2  log  x  1  1
log  x  2 x  1  1


log x 2  x  2  1
10

log x 2  x  2
  101
log  4  2  log  2
log  4  1  log  5
log 3  2  log 5
log 3  1  log 2
x 2  x  2  10
x 2  x  12  0
x  4x  3  0
x  4, x  3
The only valid solution is x
= 3.
Example #8
Radioactive Decay
 It is determined that a mummy has lost 32% of its original
carbon-14. When did the person die?
Previously, this type of problem was solved by graphing. Now we will
solve it algebraically.
A
x
P(0.5) h
We also need to know the half-life of carbon-14
which is 5730 years old.
If 32% of carbon-14 is lost, then 100% – 32% = 68%
0.68  10.5
0.68  0.5
x
5730
x
5730
Example #8
Radioactive Decay
 It is determined that a mummy has lost 32% of its original
carbon-14. When did the person die?
ln 0.68   ln 0.5
x
5730
x
 ln 0.5
ln 0.68   5730
x
ln 0.68 

ln 0.5 5730
We take the logarithm of both
sides. Then we swing the
exponent out front of the
logarithm, and solve for x.
 ln 0.68    x 
5730 

5730
 ln 0.5   5730 
5730 ln 0.68 
 3188 years ago
x
ln 0.5
Example #9
Compound Interest
 If $8000 is to be invested at 6% per year, compounded
monthly, in how many years will the investment be worth
$22,520?
 r
A  P 1  
 n
nt
22,520
 1.005 12t
8000
ln 2.815   ln 1.005 12t
12t
 0.06 
22,520  80001 

12 

22,520  8000 1.005 12t
Make sure to isolate the base of 1.005
before taking the log.
ln 2.815   12t  ln 1.005 
ln 2.815 
 12t
ln 1.005 
 ln 2.815  
t 
  12
 ln 1.005  
t  17.3 years
Example #10
Population Growth
 The population of certain species of rabbit increases exponentially. At the
beginning of a 12-month study, there were 50 rabbits, and at the end of the
study there were 164 rabbits. Find the following:
A.
Find the rate of growth.Use A 
Pa where a  1  r
x
164  50a12
After solving for a, don’t
forget to substitute 1 + r
back in for a to find r.
Here we see the growth rate
is about 10%.
164
 a12
50
a  12 3.28
a  3.28
1
12
a  1.104052
1  r  1.104052
r  0.104052
Example #10
Population Growth
 The population of certain species of rabbit increases exponentially. At the
beginning of a 12-month study, there were 50 rabbits, and at the end of the
study there were 164 rabbits. Find the following:
B.
Write the function for this population.
A  P(1  r )
t
A  Pax
To keep the function more accurate, we
will need to use a growth rate more
accurate than 10%.
A  501  0.104052 
t
A  501.104052 t
Alternatively, we can use the
value for a we found earlier.
A  50 3.28 


1
12
A  503.28 
x
12
x
Example #10
Population Growth
 The population of certain species of rabbit increases exponentially. At the
beginning of a 12-month study, there were 50 rabbits, and at the end of the
study there were 164 rabbits. Find the following:
C. When will the population reach 1000?
1000  503.2812
x
1000  501.104052 t
 1000 
t
ln 
  ln 1.104052 
 50 
ln 20   t  ln 1.104052 
ln 20 
t
ln 1.104052 
t  30.3 months
x
 1000 
12
ln 
  ln 3.28
50


x
ln 20   12
 ln 3.28
ln 20 
x

ln 3.28 12
 ln 20  
x  12

 ln 3.28 
x  30.3 months
Example #11
Population Growth
 The population of a certain bacteria culture at time t hours is
given by the function:
F (t ) 
20,000
1  20e
 3t
 How long will it take for the bacteria population to reach
8000?
The equation we will need to
solve is given by:
8000 
20,000
1  20e
 3t
Example #11
Population Growth
 How long will it take for the bacteria population to reach
8000?
8000
20,000

 3t
1
1  20e
 3t 

20,000  80001  20e 


 3t
20,000
 1  20e
8000
2.5  1  20e
1.5  20e
 3t
 3t
 3t
1.5
e
20
ln 0.075   ln e
 3t
t
ln 0.075  
3
t  3 ln 0.075 
t  7.8 hours