Transcript PowerPoint 19

MAT 150 – Class #19
Objectives
 Solve
an exponential equation by writing it in
logarithmic form
 Convert logarithms using the change of base formula
 Solve an exponential equation by using properties of
logarithms
 Solve logarithmic equations
 Solve exponential and logarithmic inequalities
Solving Exponential Equations Using Logarithmic
Forms
To solve an exponential equation using logarithmic
form:
1. Rewrite the equation with the term containing the
exponent by itself on one side.
2. Divide both sides by the coefficient of the term
containing the exponent.
3. Change the new equation to logarithmic form.
4. Solve for the variable.
Example
Solve the equation 3000  150(10 ) for t by converting it
to logarithmic form and graphically to confirm the
solution.
4t
Solution
Divide both sides of the equation by 150.
4t
3000  150(10 )
20  104t
Rewrite in logarithmic form. 4t  log20
Solve for t.
log20
t
 0.32526
4
Example (cont)
Solve the equation 3000  150(10 ) for t by converting it to
logarithmic form and graphically to confirm the solution.
4t
Solution
To solve graphically enter
3000 for y1 and 150(104t ) for y2
Example
a. Prove that the time it takes for an investment to
double its value is t  ln 2 if the interest rate is r,
r
compounded continuously.
Solution
a. S  Pert
2𝑃 = 𝑃𝑒 𝑟𝑡
2 = 𝑒 𝑟𝑡
log 𝑒 2 = 𝑟𝑡
ln 2 = 𝑟𝑡
ln2
t
r
Example (cont)
b. Suppose $2500 is invested in an account earning 6%
annual interest, compounded continuously. How long
will it take for the amount to grow to $5000?
Solution
b.
ln2
t
r
ln2
t
 11.5525
0.06
Change of Base
We can use a special formula called the change of base
formula to rewrite logarithms so that the base is 10 or e.
The general change of base formula is summarized
below.
Example
Evaluate log8 124.
Solution
log124
log8 124 
 2.318 approximately
log8
Example (cont)
b. Graph the function by changing each logarithm to a
common logarithm and then by changing the
logarithm to a natural logarithm.
y  log3 x
Solution
change to base 10
y  log3 x
log x

log3
change to base e=
ln 𝑥
ln 3
Example
If $10,000 is invested for t years at 10%, compounded
annually, the future value is given by
t
S  10,000(1.10 )
In how many years will the investment grow to $45,950?
Solution
45,950  10,000(1.10t )
4.5950  1.10t
t  log1.10 4.5950
log4.5950
t  log1.10 4.5950 
 16
log1.10
The investment will grow to $45,950 in 16 years.
SOLVING EXPONENTIAL EQUATIONS USING LOGARITHMIC
PROPERTIES
To solve an exponential equation using logarithmic
properties:
1. Rewrite the equation with a base raised to a power on
one side.
2. Take the logarithm, base e or 10, of both sides of the
equation.
3. Use a logarithmic property to remove the variable
from the exponent.
4. Solve for the variable.
Example
Solve the following exponential equations.
a. 4096  82 x
b. 6(43 x 2 )  120
Solution
a. Take log of base 10 of both sides. 4096  82 x
log4096  log82 x
Using the Power Property of Logarithms
log4096  2 x log8
Solving for x
log 4096
x
2log8
2x
Example (cont)
Solution
b. 6(43 x2 )  120
3 x 2
6(4
6
)
120

6
43 x 2  20
ln 43 x 2  ln20
(3 x  2)ln 4  ln20
ln20
3x  2 
ln 4
1  ln20

x 
 2
3  ln 4

x  1.387
Example
Solve 4log3 x  8 by converting to exponential form and
verify the solution graphically.
Solution
Divide both sides by 4: 4log3 x  8
log3 x  2
Write in exponential form: 3 2  x
1
x
9
Example
Solve 6  3ln x  12 by converting to exponential form
and verify the solution graphically.
Solution
6  3ln x  12
3ln x  6
ln x  2
x  e2
Example
Solve ln x  3  ln( x  4) by converting to exponential
form and then using algebraic methods.
Solution
ln x  3  ln( x  4)
e3 x  x  4
3  ln( x  4)  ln x
x4
3  ln
x
x4
3
e 
x
e3 x  x  4
x(e3  1)  4
4
x 3
 0.21
e 1
Example
After the end of an advertising campaign, the daily sales
of Genapet fell rapidly, with daily sales given by
S = 3200e-0.08x dollars, where x is the number of days
from the end of the campaign. For how many days after
the campaign ended were sales at least $1980?
Solution
3200e 0.08 x  1980
e 0.08 x  0.61875
0.08 x
ln e
 ln0.61875
0.08x  0.4801
x 6
Assignment
Pg. 350-354
#3-7 odd
#15-33 odd
#47
#66