8.6 Solving Exponential and Logarithmic Equations

p. 501

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?

• 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

Exponential Equations

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

Your turn!

• 24x = 32x-1

• 24x = (25)x-1

• 4x = 5x-5

• 5 = x

Be sure to check your answer!!!

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 = ≈ 2.8072log

7log

Use log2 because the x is on the 2 and log22=1

4x = 15• log44x = log415

• x = log415 = log15/log4

• ≈ 1.95

Use change of base to solve

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

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

Newton’s Law of Cooling

• The temperature T of a cooling substance @ time t (in minutes) is:

•T = (T0 – TR) e-rt + TR

• T0= initial temperature

• TR= room temperature

• r = constant cooling rate of the substance

• 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 =.046. How long will it take to cool the stew to a serving temp. of 100°?

• T0 = 212, TR = 70, T = 100 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?

• ln .221 ≈ ln e-.046t (take the ln of both sides)

• ln .221 ≈ -.046t

• -1.556 ≈ -.046t

• 33.8 ≈ t

• about 34 minutes to cool!

Cooling cont.

• 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.

Assignment 8.6

Page 505, 25-40, 62-65

Solving Logarithmic Equations 8.6

Day 2

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

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

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

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

log5x + log(x-1)=2• log (5x)(x-1) = 2 (product property)

• log (5x2 – 5x) = 2 (use definition)

• 5x2−5x = 102

• 5x2 - 5x = 100

• x2 – x - 20 = 0 (subtract 100 and divide by 5)

• (x-5)(x+4) = 0 x=5, x=-4• graph and you’ll see 5=x is the only solution

2

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

Assignment 8.6 day 2

• p. 505, 43-60, skip 51 & 52