1. 8.6 Solving Exponential and Logarithmic Equations
How do you use logs to solve an exponential equation?
When is it easiest to use the definition of logs?
Do you ever get a negative answer for logs?

2. Exponential Equations
One way to solve exponential equations is to use the property that if 2 powers w/ the same base are equal, then their exponents are equal.
For b>0 & b≠1 if bx = by, then x=y

3. Solve by equating exponents
43x = 8x+1
(22)3x = (23)x+1 rewrite w/ same base
26x = 23x+3
6x = 3x+3
x = 1
Check → 43*1 = 81+1
64 = 64

4. 24x = 32x-1 24x = (25)x-1 4x = 5x-5 5 = x Your turn!
Be sure to check your answer!!!

5. When you can’t rewrite using the same base, you can solve by taking a log of both sides
2x = 7
log22x = log27
x = log27
x = ≈ 2.807
Use log2 because the x is on the 2 and log22=1

6. 4x = 15 log44x = log415 x = log415 = log15/log4 ≈ 1.95
Use change of base to solve

7. 102x-3+4 = 21 -4 -4 102x-3 = 17 log10102x-3 = log1017 2x-3 = log 17
102x-3 = 17
log10102x-3 = log1017
2x-3 = log 17
2x = 3 + log17
x = ½(3 + log17)
≈ 2.115

8. 5x+2 + 3 = 25 5x+2 = 22 log55x+2 = log522 x+2 = log522
= (log22/log5) – 2
≈ -.079

9. Newton’s Law of Cooling
The temperature T of a cooling time t (in minutes) is:
T = (T0 – TR) e-rt + TR
T0= initial temperature
TR= room temperature
r = constant cooling rate of the substance

10. You’re cooking stew. When you take it off the stove the temp. is 212°F
You’re cooking stew. When you take it off the stove the temp. is 212°F. The room temp. is 70°F and the cooling rate of the stew is r = How long will it take to cool the stew to a serving temp. of 100°?

11. So solve: 100 = (212 – 70)e-.046t +70 30 = 142e-.046t (subtract 70)
T0 = 212, TR = 70, T = r = .046
So solve:
100 = (212 – 70)e-.046t +70
30 = 142e-.046t (subtract 70)
.221 ≈ e-.046t (divide by 142)
How do you get the variable out of the exponent?

12. ln .221 ≈ ln e-.046t (take the ln of both sides) ln .221 ≈ -.046t
Cooling cont.
ln .221 ≈ ln e-.046t (take the ln of both sides)
ln .221 ≈ -.046t
≈ -.046t
33.8 ≈ t
about 34 minutes to cool!

13. How do you use logs to solve an exponential equation?
Expand the logs to bring the exponent x down and solve for x.
When is it easiest to use the definition of logs?
When you have log information on the left equal to a number on the right.
Do you ever get a negative answer for logs?
Never! Logs are always positive.

14. Assignment 8.6
Page 505, 25-40,

15. Solving Logarithmic Equations 8.6 Day 2

16. If logbx = logby, then x = y
Solving Log Equations
To solve use the property for logs w/ the same base:
+ #’s b,x,y & b≠1
If logbx = logby, then x = y

17. 5x – 1 = x + 7 5x = x + 8 4x = 8 x = 2 and check
log3(5x-1) = log3(x+7)
5x – 1 = x + 7
5x = x + 8
4x = 8
x = 2 and check
log3(5*2-1) = log3(2+7)
log39 = log39

18. When you can’t rewrite both sides as logs w/ the same base exponentiate each side
b>0 & b≠1
if x = y, then bx = by

19. 3x+1 = 25 x = 8 and check log5(3x + 1) = 2
52 = (3x+1) (use definition)
3x+1 = 25
x = 8 and check
Because the domain of log functions doesn’t include all reals, you should check for extraneous solutions

20. log5x + log(x-1)=2
log (5x)(x-1) = (product property)
log (5x2 – 5x) = 2 (use definition)
5x2−5x = 102
5x2 - 5x = 100
x2 – x - 20 = (subtract 100 and divide by 5)
(x-5)(x+4) = x=5, x=-4
graph and you’ll see 5=x is the only solution 2

21. One More! log2x + log2(x-7) = 3
log2x(x-7) = 3
log2 (x2- 7x) = 3
x2−7x = 23
x2 – 7x = 8
x2 – 7x – 8 = 0
(x-8)(x+1)=0
x=8 x= -1 2

22. Assignment 8.6 day 2
p. 505, 43-60, skip 51 & 52