Transcript 7.6: Solving Exponential and Logarithmic Equations

Solving Exponential and Logarithmic Equations
Objectives:
1. To solve exponential equations
2. To solve logarithmic equations
Objective 1
You will be able to solve exponential equations
Exercise 1
If 2x = 23, then what is the value of x?
Equality of Exponentials
Property of Equality of Exponential
Equations
If 𝑏 is a positive real number not equal to
1, then 𝑏𝑥 = 𝑏𝑦 iff 𝑥 = 𝑦.
2𝑥 = 23
iff 𝑥 = 3
This means that one way to solve
exponential equations is to make the bases
equal so that the exponents are equal.
Exercise 2
Solve
3𝑥
=
1 𝑥+3
9
Exercise 3
Solve 9𝑥 = 25.
Taking the Log of Both Sides
Sometimes you can’t rewrite your equation so that
the bases are the same (or maybe you just don’t
want to). In that case, try one of the following:
1. Take the log
of both sides
2. Rewrite the equation
in logarithmic form
– Both of these are really the same thing
– You may have to use the change of base formula
Exercise 3
9𝑥 = 25
log 9 9𝑥 = log 9 25
𝑥 = log 9 25
log 25
𝑥=
log 9
𝑥 ≈ 1.465
Method 1:
Take the log9 of both sides. The
base of the exponential becomes
the base of the log.
Use the Inverse Property to get 𝑥 on
the left.
Use the Change of Base formula.
Exercise 3
9𝑥 = 25
Method 1e:
ln 9𝑥 = ln 25
Just take the natural or the common
log of both sides. Pick your favorite.
𝑥 ∙ ln 9 = ln 25
Use the Power Property to get 𝑥 out
front.
ln 25
𝑥=
ln 9
𝑥 ≈ 1.465
Divide by ln 9. Notice this does the
Base Changing for you.
Exercise 3
9𝑥 = 25
log 9 25 = 𝑥
log 25
𝑥=
log 9
𝑥 ≈ 1.465
Method 2:
Write the original equation in
logarithmic form
Use the Change of Base formula.
Exercise 4
Solve −3𝑒2𝑥 + 16 = 5.
Exercise 5
Solve each equation.
1. 8x – 1 = 323x – 2
2.
2x = 5
3.
79x = 15
4.
4e−0.3x – 7 = 13
Exercise 5
Solve each equation.
1. 8x – 1 = 323x – 2
2. 2x = 5
Exercise 5
Solve each equation.
3. 79x = 15
4. 4e−0.3x – 7 = 13
Newton’s Law of Cooling
For a cooling substance (like
a cup of hot chocolate) with
an initial temperature of T0,
the temperature T after t
minutes can be modeled
by
T = (T0 – TR)e−rt + TR,
where TR is the
surrounding temperature
and r is the cooling rate.
Exercise 6a
You warm a mug of hot chocolate
to a temperature of 90°C. You
leave the warm beverage on
the counter while you go to
tend to your wailing baby. If
the hot chocolate in the mug
has a cooling rate of 0.145 and
the temperature of the room is
20°C, how many minutes will it
take for the hot chocolate to
cool to a temperature of 30°C?
Exercise 6b
Explain how to find the cooling rate of a
substance.
Exercise 7
A new car costs $25,000. The value of the
car decreases by 15% each year. Write an
exponential decay model giving the car’s
value y (in dollars) after t years. In how
many years with the value of the car be ½
the original value?
Exercise 8
According to NATO, a movie ticket
to see Star Wars (Episode IV: A
New Hope) in 1977 was an
average price of $2.23. In 1997
when Star Wars was re-released
to theaters in anticipation of the
upcoming prequel trilogy, ticket
prices averaged $4.59.
Use the growth model y = Pert to
find the annual growth rate from
1977 to 1997.
Objective 2
You will be able to solve a
logarithmic equation
Exercise 9
If log2 x = log2 25, then what is the value of
x?
Equality of Logarithmic Equations
Property of Equality of Logarithmic
Equations
If 𝑏, 𝑥, and 𝑦 are positive real numbers
with 𝑏 ≠ 1, then log𝑏 𝑥 = log𝑏 𝑦 iff 𝑥 = 𝑦.
log2 𝑥 =
log2 25 iff
𝑥 = 25
This means that one way to solve
logarithmic equations is to make the bases
equal so that what you are taking the logs
of are equal.
Exercise 10
Solve log4 (2𝑥 + 8) = log4 (6𝑥 – 12).
Exponentiating
Since 𝑥 = 𝑦 means 𝑏𝑥 = 𝑏𝑦, we can
exponentiate both sides of an equation
using the same base.
Notice the two sides of
the equation become
the exponents
Exponentiate means to
raise a quantity to a
power
Essentially rewriting a
logarithmic equation as
an exponential equation
Check for extraneous
solutions
Exercise 11
Solve log7 (3x – 2) = 2.
Method 1:
log 7 3𝑥 − 2 = 2
7log7
3𝑥−2
= 72
Exponentiate using 7 as the base.
3𝑥 − 2 = 49
Use the Inverse Property to simplify.
3𝑥 = 51
Solve for 𝑥.
𝑥 = 17
Exercise 11
Solve log7 (3x – 2) = 2.
Method 2:
log 7 3𝑥 − 2 = 2
72 = 3𝑥 − 2
3𝑥 − 2 = 49
3𝑥 = 51
𝑥 = 17
Write in exponential form.
Solve for 𝑥.
Exercise 12
Solve log6 3x + log6 (x – 4) = 2.
Exercise 13
Solve each equation.
1. ln (7x – 4) = ln (2x + 11)
2. log2 (x – 6) = 5
3. log 5x + log (x – 1) = 2
4. log4 (x + 12) + log4 x = 3
Exercise 13
Solve each equation.
1. ln (7x – 4) = ln (2x + 11)
2. log2 (x – 6) = 5
Exercise 13
Solve each equation.
3. log 5x + log (x – 1) = 2
4. log4 (x + 12) + log4 x = 3
Special Cases: Quadratic
Sometimes we want to solve an equation that looks
kind of like a quadratic. Well, just treat it like one.
𝑒 2𝑥 − 7𝑒 𝑥 + 12 = 0
𝑒𝑥 − 3 𝑒𝑥 − 4 = 0
𝑒𝑥 − 3 = 0
𝑒𝑥 − 4 = 0
𝑒𝑥 = 3
𝑒𝑥 = 4
𝑥 = ln 3
𝑥 ≈ 1.099
𝑥 = ln 4
𝑥 ≈ 1.386
Let 𝑘 = 𝑒 𝑥
So 𝑘 2 = 𝑒 2𝑥
𝑘 2 − 7𝑘 2 + 12 = 0
𝑘−3 𝑘−4 =0
Special Cases: Different Base
What if the exponential has different bases? Just pick one
to take the log of, the other one will become an exponent.
Or just take the natural log!
34𝑥−5 = 72𝑥
ln 34𝑥−5 = ln 72𝑥
4𝑥 − 5 ln 3 = 2𝑥 ∙ ln 7
4𝑥 ∙ ln 3 − 5 ln 3 = 2𝑥 ∙ ln 7
4𝑥 ∙ ln 3 − 2𝑥 ∙ ln 7 = 5 ln 3
𝑥 4ln 3 − 2ln 7 = 5 ln 3
5 ln 3
𝑥=
4ln 3 − 2ln 7
𝑥 ≈ 10.929
Special Cases: Different Base
What if the logarithms have different bases, one of which is
a power of the other? Use a change of base—keep the
smaller, get rid of the larger.
log 3 2𝑥 = log 9 13𝑥 − 3
log 3 13𝑥 − 3
log 3 2𝑥 =
log 3 9
log 3 13𝑥 − 3
log 3 2𝑥 =
2
2 ∙ log 3 2𝑥 = log 3 13𝑥 − 3
log 3 2𝑥
2
= log 3 13𝑥 − 3
4𝑥 2 = 13𝑥 − 3
4𝑥 2 − 13𝑥 + 3 = 0
4𝑥 − 1 𝑥 − 3 = 0
𝑥 = 1/4
𝑥=3
Solving Exponential and Logarithmic Equations
Objectives:
1. To solve exponential
equations
2. To solve logarithmic
equations