1. WARMUP Lesson 7.6, For use with pages 515-522
1. Write log3(2x – 7) = 4 in exponential form.
ANSWER
34 = 2x – 7
Write 8x = 30 in logarithmic form.
Hint: reverse loop swoop doop
ANSWER
log8 30 = x
Solve for x.
ANSWER 3 2
3. 100x = 1000
ANSWER 1
125
4. log5x = –3
x2 – 7x – 60 = 0
Hint: Use the quadratic formula.
ANSWER
–5, 12

2. 7.6 Notes - Solve Exponential and Log Equations

3. Objective - To solve exponential and logarithmic
Rewrite each base as a power of 2.
Distribute.
Powers equal to each other.
Objective - To solve exponential and logarithmic
equations.

5. Take the log3 of each side.

6. Take the log5 of each side.

7. If logb x = logb y if and only if x = y.

8. Loop Swoop Doop

9. If x = y, then bx = by. You do this one
You probably did loop-swoop-doop, but here is another approach.

10. Check! Remember, you can’t take the log of a negative!10
Check! Remember, you can’t take the log of a negative!

11. Solve by equating exponents
EXAMPLE 1
Solve by equating exponents
Solve 4 = x 1 2
x – 3
SOLUTION
4 = x 1 2
x – 3
Write original equation.
Rewrite 4 and as powers with base 2. 1 2
(2 ) = (2 ) 2
x – 3 x
– 1
2 = 2
2 x
– x + 3
Power of a power property
2x = –x + 3
Property of equality for exponential equations
x = 1
Solve for x.
The solution is 1.
ANSWER

12. Solve by equating exponents
EXAMPLE 1
Solve by equating exponents
Check: Check the solution by substituting it into the original equation.
1 – 3
4 = 1 ? 2
Substitute 1 for x.
4 = ? 1 2
– 2
Simplify.
4 = 4
Solution checks.

13. GUIDED PRACTICE for Example 1 Solve the equation. 3. 81 = 1 3=
3 – x 1 3
5x – 6
= 27 2x
x – 1
SOLUTION –3
SOLUTION –6
= 1000
7x + 1
3x – 2
– 8 5
SOLUTION

14. Take a logarithm of each side
EXAMPLE 2
Take a logarithm of each side
Solve 4 = 11. x
SOLUTION
4 = 11 x
Write original equation.
log = log 11 x 4
Take log of each side. 4
log 4
x =
log b = x b x
x =
log 11
log 4
Change-of-base formula x
Use a calculator.
The solution is about Check this in the original equation.
ANSWER

15. EXAMPLE 3
Use an exponential model
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 = and it is 80°F outside,how long (in minutes) do you have to wait until you can continue driving?
Cars

16. Use an exponential model
EXAMPLE 3
Use an exponential model
SOLUTION
T = ( T – T )e T R
– rt
Newton’s law of cooling
230 = (280 – 80)e
–0.0048t
Substitute for T, T , T , and r. R
150 = 200e
–0.0048t
Subtract 80 from each side.
0.75 = e
–0.0048t
Divide each side by 200.
ln 0.75 = ln e
–0.0048t
Take natural log of each side.
– –0.0048t
In e = log e = x e x t
Divide each side by –

17. EXAMPLE 3
Use an exponential model
You have to wait about 60 minutes until you can continue driving.
ANSWER

18. GUIDED PRACTICE for Examples 2 and 3 Solve the equation. 4. 2 = 5
= 5 x
SOLUTION
about 2.32
= 15 9x
SOLUTION
about
e –7 = 13
–0.3x
SOLUTION
about –5.365

19. Solve a logarithmic equation
EXAMPLE 4
Solve a logarithmic equation
Solve log (4x – 7) = log (x + 5). 5
SOLUTION
log (4x – 7) = log (x + 5). 5
Write original equation.
4x – 7 = x + 5
Property of equality for logarithmic equations
3x – 7 = 5
Subtract x from each side.
3x = 12
Add 7 to each side.
x = 4
Divide each side by 3.
The solution is 4.
ANSWER

20. Solve a logarithmic equation
EXAMPLE 4
Solve a logarithmic equation
Check: Check the solution by substituting it into the original equation.
(4x – 7) = (x – 5)
log 5
Write original equation.
(4 4 – 7) = (4 + 5) ?
log 5
Substitute 4 for x.
9 =
log 5
Solution checks.

21. Exponentiate each side of an equation
EXAMPLE 5
Exponentiate each side of an equation
Solve (5x – 1)= 3
log 4
SOLUTION
(5x – 1)=
(5x – 1)= 3
log 4
Write original equation.
4log4(5x – 1) = 4 3
Exponentiate each side using base 4.
b = x
log b x
5x – 1 = 64
5x = 65
Add 1 to each side.
x = 13
Divide each side by 5.
The solution is 13.
ANSWER

22. Exponentiate each side of an equation
EXAMPLE 5
Exponentiate each side of an equation
log 4
(5x – 1) = (5 13 – 1) =
Check:
log 4
Because 4 = 64, = 3. 3

23. Standardized Test Practice
EXAMPLE 6
Standardized Test Practice
SOLUTION
log 2x + log (x – 5) = 2
Write original equation.
log [2x(x – 5)] = 2
Product property of logarithms
= 10
log 2
[2x(x – 5)]
Exponentiate each side using base 10.
2x(x – 5) = 100
Distributive property

24. Standardized Test Practice
EXAMPLE 6
Standardized Test Practice
2x – 10x = 100 2
b = x
log b x
2x – 10x – 100 = 0 2
Write in standard form.
x – 5x – 50 = 0 2
Divide each side by 2.
(x – 10)(x + 5) = 0
Factor.
x = 10 or x = – 5
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.

25. 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
Because log (–10) is not defined, –5 is not a solution.
2 = 2
So, 10 is a solution.

26. 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.
The correct answer is C.
ANSWER

27. Solve the equation. Check for extraneous solutions.
GUIDED PRACTICE
for Examples 4, 5 and 6
Solve the equation. Check for extraneous solutions.
7. ln (7x – 4) = ln (2x + 11)
9. log 5x + log (x – 1) = 2
SOLUTION 5
SOLUTION 3
8. log (x – 6) = 5 2
log (x + 12) + log x =3 4
SOLUTION 4
SOLUTION 38

28. EXAMPLE 7
Use a logarithmic model
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
Astronomy
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?

29. Use a logarithmic model
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
Divide each side by 5.
Exponentiate each side using base 10.
10 = 10 2
Log D
100 = D
Simplify.
The diameter is 100 millimeters.
ANSWER

30. GUIDED PRACTICE
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.

31. 7.6 Assignment
Hint #3: Quadratic Formula

32. WARMUP Daily Homework Quiz For use after Lesson 7.6 Solve. ANSWER 6 5
x = 125 –x + 2
x = 5
ANSWER
about 0.77
3. log7 (5x – 8) = log7 (2x + 19)
ANSWER 9
4. log3 (5x + 1) = 4 16
ANSWER
5. log5 5x + log5 (x – 4) = 2 5
ANSWER
Bonus. Boiling water has a temperature of 212° F. Water has a cooling rate of r = 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.
ANSWER
about 13 min

33. Chapter 7 Test Review Assignment p.541: 17-20 all, 24-34 all (16 Q’s)
p.543: all, 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)

34. Answers to the Review (p.541) 16 Q’s-3
-1/3
6.86mm, 5.39mm
log83 + log8 x + log8 y
ln10 + 3ln x + ln y
log8 – 4 log y
ln 3 + ln y – 5 ln x
28. ln 3 + ln y – 5 ln x
29. log7 384
30. ln(12/x2)
31. ln36
33.7

35. Answers to the Test (p.543) 13 Q’s2 -5
16. ln(49/64)
17. log496
log(5x/9)
2.431
1.750
1.732
0.874
104 2
28. $