8.6Solving Exponential and Logarithmic Equations

p. 501

• 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

4x = 15• log44x = log415

• x = log415 = log15/log4

• ≈ 1.95

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.

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

• 5log5

(3x+1) = 52

• 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

• 10log5x -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

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One More!

log2x + log2(x-7) = 3• log2x(x-7) = 3• log2 (x2- 7x) = 3• 2log

2x -7x = 32

• x2 – 7x = 8• x2 – 7x – 8 = 0• (x-8)(x+1)=0• x=8 x= -1

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Assignment