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WARMUP
Lesson 7.6, For use with pages 515-522
1. Write log3(2x – 7) = 4 in exponential form.
34 = 2x – 7
ANSWER
2. Write 8x = 30 in logarithmic form. ANSWER
Hint: reverse loop swoop doop
Solve for x.
3
3. 100x = 1000
ANSWER
2
4. log5x = –3
ANSWER
5. x2 – 7x – 60 = 0
Hint: Use the
quadratic formula.
ANSWER
1
125
–5, 12
log8 30 = x
7.6 Notes - Solve Exponential
and Log Equations
https://www.youtube.com/watch?v=X
jzBpJ1XYwc
Solve 4  16
5x
2 
2 5x
10x
2
7 x 6
 2
2

Rewrite each base as a power of 2.
4 7 x 6
28 x24
Distribute.
Powers equal to each other.
10x  28x  24
4
x
3
Objective - To solve exponential and logarithmic
equations.
Solve 25
5
2x1
 625
2 2x 1
5
4 x2
5
5
4 4 x 
16 x
4x  2  16x
1
x
6
4x
Solve 3  11
x
Take the log3 of each side.
log 3 3  log 311
x
x  log 311
log 11 1.0414

x
log 3 0.4771
x  2.1828
Solve 5  27
x
Take the log5 of each side.
log 5 5  log 5 27
x
x  log 5 27
log 27 1.4314

x
log 5 0.6990
x  2.0478
Solve log7  4 x  9  log7  2 x  5
If logb x = logb y if and only if x = y.
4x  9  2x  5
2x  4
x2
Solve log6  4 x  9   3
Loop Swoop Doop
6  4x  9
3
216  4x  9
207
x
4
Solve log3  2 x  5  2
If x = y, then bx = by.
log 3 2 x  5 
3
3
2x  5  9
x  2
You do
this one
2
You probably
did loop-swoopdoop, but here
is another
approach.
Solve logx  log(x  3)  1
log  x(x  3)  1
log10 x  3 x   1
2
10  x  3x
2
x  3x  10  0
(x  5)(x  2)  0
1
2
x  5 x  2
Check! Remember, you
can’t take the log of a
negative!
EXAMPLE 1
Solve by equating exponents
x
Solve 4 = 1
2
x–3
SOLUTION
1
4 = 2
x–3
x
–1 x–3
2 x
(2 ) = (2 )
22 x = 2– x + 3
2x = –x + 3
x=1
ANSWER
Write original equation.
1
Rewrite 4 and 2 as powers with
base 2.
Power of a power property
Property of equality for
exponential equations
Solve for x.
The solution is 1.
EXAMPLE 1
Solve by equating exponents
Check: Check the solution by substituting it into the
original equation.
41
1
= 2
?
?
4 =
4=4
1
2
1–3
Substitute 1 for x.
–2
Simplify.
Solution checks.
GUIDED PRACTICE
for Example 1
Solve the equation.
2x
1. 9
= 27
SOLUTION
2.
x–1
–3
1007x + 1 = 10003x – 2
SOLUTION
–8
5
3.
3–x
81
1
= 3
SOLUTION
5x – 6
–6
EXAMPLE 2
Take a logarithm of each side
x
Solve 4 = 11.
SOLUTION
x
Write original equation.
4 = 11
x
log 44 = log 411
x = log 4 11
log 11
x = log 4
x
ANSWER
1.73
Take log of each side.
4
logb b x = x
Change-of-base formula
Use a calculator.
The solution is about 1.73. Check this in
the original equation.
EXAMPLE 3
Use an exponential model
Cars You are driving on a hot day when your car
overheats and stops running. It overheats at 280°F
and can be driven again at 230°F. If r = 0.0048 and it
is 80°F outside,how long (in minutes) do you have
to wait until you can continue driving?
EXAMPLE 3
Use an exponential model
SOLUTION
T = ( T°– TR )e – rt + T R
Newton’s law of cooling
230 = (280 – 80)e –0.0048t + 80 Substitute for T, T , TR, and r.
°
150 = 200e–0.0048t
Subtract 80 from each side.
0.75 = e–0.0048t
ln 0.75 = ln e –0.0048t
–0.2877
60
Divide each side by 200.
Take natural log of each side.
–0.0048t
In ex = loge ex = x
t
Divide each side by –0.0048.
EXAMPLE 3
Use an exponential model
ANSWER
You have to wait about 60 minutes until you can
continue driving.
GUIDED PRACTICE
for Examples 2 and 3
Solve the equation.
4. 2 x = 5
SOLUTION
5.
7 9x = 15
SOLUTION
6.
about 2.32
about 0.155
4e –0.3x –7 = 13
SOLUTION
about –5.365
EXAMPLE 4
Solve a logarithmic equation
Solve log 5 (4x – 7) = log 5(x + 5).
SOLUTION
log 5 (4x – 7) = log 5(x + 5).
4x – 7 = x + 5
3x – 7 = 5
ANSWER
Property of equality for
logarithmic equations
Subtract x from each side.
3x = 12
x=4
Write original equation.
Add 7 to each side.
Divide each side by 3.
The solution is 4.
EXAMPLE 4
Solve a logarithmic equation
Check: Check the solution by substituting it into the
original equation.
log 5(4x – 7) = log 5(x – 5)
log 5(4 4 – 7) =? log (4 + 5)
5
log 5 9 = log 5 9
Write original equation.
Substitute 4 for x.
Solution checks.
EXAMPLE 5
Exponentiate each side of an equation
Solve log 4 (5x – 1)= 3
SOLUTION
log 4 (5x – 1)= 3
4log4(5x – 1) =
4
3
5x – 1 = 64
5x = 65
x = 13
ANSWER
Write original equation.
Exponentiate each side using
base 4.
blogbx = x
Add 1 to each side.
Divide each side by 5.
The solution is 13.
EXAMPLE 5
Exponentiate each side of an equation
Check:log 4 (5x – 1) = log 4(5 13 – 1) = log 4 64
3
Because 4 = 64, log4 64= 3.
EXAMPLE 6
Standardized Test Practice
SOLUTION
log 2x + log (x – 5) = 2
log [2x(x – 5)] = 2
2
log [2x(x – 5)]
10
= 10
2x(x – 5) = 100
Write original equation.
Product property of logarithms
Exponentiate each side using
base 10.
Distributive property
EXAMPLE 6
Standardized Test Practice
2
2x – 10x = 100
2
2x – 10x – 100 = 0
x 2 – 5x – 50 = 0
(x – 10)(x + 5) = 0
x = 10 or x = – 5
blog bx = x
Write in standard form.
Divide each side by 2.
Factor.
Zero product property
Check: Check the apparent solutions 10 and –5 using
algebra or a graph.
Algebra: Substitute 10 and –5 for x in the original
equation.
EXAMPLE 6
Standardized Test Practice
log 2x + log (x – 5) = 2
log 2x + log (x – 5) = 2
log (2 10) + log (10 – 5) = 2
log [2(–5)] + log (–5 – 5) = 2
log 20 + log 5 = 2
log (–10) + log (–10) = 2
log 100 = 2
2=2
So, 10 is a solution.
Because log (–10) is not
defined, –5 is not a
solution.
EXAMPLE 6
Standardized Test Practice
Graph: Graph y = log 2x + log (x – 5)
and y = 2 in the same coordinate
plane. The graphs intersect only
once, when x = 10. So, 10 is the
only solution.
ANSWER
The correct answer is C.
GUIDED PRACTICE
for Examples 4, 5 and 6
Solve the equation. Check for extraneous solutions.
7.
ln (7x – 4) = ln (2x + 11)
SOLUTION
8.
3
log 2 (x – 6) = 5
SOLUTION
38
9.
log 5x + log (x – 1) = 2
SOLUTION
10.
5
log 4 (x + 12) + log 4x =3
SOLUTION
4
EXAMPLE 7
Use a logarithmic model
Astronomy The apparent magnitude of a star is a
measure of the brightness of the star as it appears to
observers on Earth. The apparent magnitude M of the
dimmest star that can be seen with a telescope is
given by the function
M = 5 log D + 2
where D is the diameter (in millimeters) of the
telescope’s objective lens. If a telescope can reveal
stars with a magnitude of 12, what is the diameter of
its objective lens?
EXAMPLE 7
Use a logarithmic model
SOLUTION
M = 5 log D + 2
Write original equation.
12 = 5 log D + 2
Substitute 12 for M.
10 = 5 log D
Subtract 2 from each side.
2 = log D
2
10 = 10 Log D
100 = D
ANSWER
Divide each side by 5.
Exponentiate each side using
base 10.
Simplify.
The diameter is 100 millimeters.
GUIDED PRACTICE
11.
for Example 7
WHAT IF? Use the information from Example 7 to
find the diameter of the objective lens of a telescope
that can reveal stars with a magnitude of 7.
SOLUTION
The diameter is 10 millimeters.
7.6 Assignment
Hint #3: Quadratic Formula
WARMUP
Daily Homework Quiz
For use after Lesson 7.6
Solve.
1. 25x = 125 –x + 2
2. 8x = 5
ANSWER
ANSWER
6
5
about 0.77
3. log7 (5x – 8) = log7 (2x + 19)
4. log3 (5x + 1) = 4
ANSWER
5. log5 5x + log5 (x – 4) = 2
ANSWER
9
16
ANSWER
5
Bonus. Boiling water has a temperature of 212° F.
Water has a cooling rate of r = 0.042. Use the
formula T = (T0 – TR)e –rt + TR to find the number of
minutes t it will take for the water to cool to a
temperature of 80°F if the room temperature is 72°F.
about 13 min
ANSWER
Chapter 7 Test
Review Assignment
p.541: 17-20 all, 24-34 all (16 Q’s)
Chapter Test
p.543: 10-12 all, 16-24 all, 28 (A = Pert)
(13 Q’s)
Do as much as you can without a calculator.
Use your calculator to check (if not in the back)
Answers to the Review (p.541) 16 Q’s
17.5
18.0
19.-3
20.-1/3
24.6.86mm, 5.39mm
25.log83 + log8 x + log8 y
26.ln10 + 3ln x + ln y
27.log8 – 4 log y
28.ln 3 + ln y – 5 ln x
28. ln 3 + ln y – 5 ln x
29. log7 384
30. ln(12/x2)
31. ln36
32. 2.153
33.7
34. 3.592
Answers to the Test (p.543) 13 Q’s
10.2
11.-5
12.0
16. ln(49/64)
17. log496
18.log(5x/9)
19.2.431
20.1.750
21.1.732
22.0.874
23.104
24.2
28. $3307.82