Solving Exponential and Logarithmic Equations Solving when the exponent has a variable.

Rules of exponents Multiply……add Divide………subtract Power to a power….multiply

Radicals become Fractional exponents:

Practice with exponents See textbook pg. 292

Radical expressions Express the when coverting a radical Expression with a fractional exponent. Example: 1. What is the power? 2. what is the root?

Fractional exponents Express as Now evaluate: = 32

Practice: See page 296 in textbook.do 3-6 297 20, 21 46-48, 58, 68,69 297 20, 21 46-48, 58, 68,69 Do these:

Do now: Express with fractional exponents: Solve:

Solving when the base is a variable: Solve the previous problem:

Procedure: Isolate the exponential, raise to the reciprocal power, and solve:

Procedure: Isolate the exponential, raise to the reciprocal power, and solve:

Example: Isolate first!

Example: Isolate first!

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. if bx = by, then x=y For b>0 & b≠1

base 2, base 3,base 5 numbers: 2 3 5 4 9 25 8 27 125 16 81 32 64

Solve by equating exponents 43x = 8x+1 But 4 = 22 and 8 = 23 (22)3x = (23)x+1 rewrite w/ same base

Solve by equating exponents 43x = 8x+1 (22)3x = (23)x+1 rewrite w/ same base 26x = 23x+3 power to a power: multiply 6x = 3x+3 set exponents equal x = 1 Check → 43*1 = 81+1 64 = 64

Procedure: Copy these steps 1. decide like base 2. rewrite one or both numbers as a power of the base 3. simplify the exponents 4. set exponents equal and solve 5. be sure to check!

Your turn! 24x = 32x-1

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

Hand this one in: Solve and check:

Hand this one in: Solve and check:

Using fractions Solve for x:

Using fractions Solve for x:

Using fractions Solve for x:

Using fractions Solve for x:

Solving with quadratic exponents Solve and check for both solutions:

Solving with quadratic exponents Solve and check for both solutions:

Review questions Add to an index card: (remember )

Review questions Add to an index card:

Exponential growth/decay rate formula: Growth: A=a(1+r)t Decay: A=a(1-r)t Which becomes….. a is the initial value b is used to find the growth or decay rate. x or t is the time If b >1 then b - 1 is the growth rate If b < 1 then 1 – b is the decay rate Note: the rate is always in decimal form.

Graphs of exponentials Growth and decay: Growth b>1 Decay b<1

Interest rate Use the formula to determine the amount of money you will have after 3 years if you invest 100 at a rate of 8%. A=a0(1+r)t

Interest rate Use the formula to determine the amount of money you will have after 3 years if you invest 100 at a rate of 8%.

Exponential examples Y = 300(.95)x What is the initial value? What is the rate of decay? How much will there be in 10 years? Y = 20(1.3)t What is the initial value What is the growth rate? How much will it be in 5 years?

Exponential examples Y = 300(.95)x What is the initial value? 300 What is the rate of decay? 1-.95=.05=5% How much will there be in 10 years? 179.62 Y = 20(1.3)t What is the initial value 20 What is the growth rate? 1.3-1=.3=30% How much will it be in 5 years? 20(1.30)5=74.26

COMPOUND INTEREST FORMULA annual interest rate (as a decimal) Principal (amount at start) time (in years) amount at the end number of times per year that interest in compounded

Calculate how much you will have if You invest 100 at 8% compounded Quarterly after 3 years.

Calculate how much you will have if You invest 100 at 8% compounded Quarterly after 3 years.

Compound or CONTINUOUS growth: The rate is changed to a decimal if not already. Ex: population grows continuously at a rate of 2% in Allentown. If Allentown has 10,000 people today, how many will it have 5 years From now?

Compound or CONTINUOUS growth: Ex: population grows continuously at a rate of 2% in Allentown. If Allentown has 10,000 people today, how many will it have 5 years From now?

How much will a 1000 investment be Worth in 10 years if it is compounded continuously at 2.5%? If a substance decays continually at a rate of -0.05 an hour, How much will be left of a 20 gram substance in 6 hours?

How much will a 1000 investment be Worth in 10 years if it is compounded continuously At 2.5%? If a substance decays continually at a rate of -0.05 an hour, How much will be left of a 20 gram substance in 6 hours?

2x = 7 log 2x = log 7 x log 2 = log 7 x = ≈ 2.807 When you can’t rewrite using the same base, you can solve by taking a log of both sides 2x = 7 log 2x = log 7 x log 2 = log 7 x = ≈ 2.807

4x = 15 “log” both sides & solve

4x = 15 log 4x = log 15 x log 4 = log15 x= log 15/log 4 ≈ 1.95

10(2x-3) +4 = 21 -4 -4 102x-3 = 17

10(2x-3) +4 = 21 -4 -4 102x-3 = 17 log102x-3 = log17 -4 -4 102x-3 = 17 log102x-3 = log17 (2x-3)(log 10) = log 17 2x – 3 = log17/log 10 2x =(3 +1.2304) x= ≈ 2.115

5x+2 + 3 = 25 isolate and solve:

5x+2 + 3 = 25 5x+2 = 22 log5x+2 = log 22 (x+2) log 5 = log 22 ≈ -.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 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°?

So solve: 100 = (212 – 70)e-.046t +70 30 = 142e-.046t (subtract 70) T0 = 212, TR = 70, T = 100 r = .046 So solve: 100 = (212 – 70)e-.046t +70 30 = 142e-.046t (subtract 70) .211 ≈ e-.046t (divide by 142) How do you get the variable out of the exponent?

ln .211 ≈ ln e-.046t (take the ln of both sides) ln .211 ≈ -.046t ln e Cooling cont. ln .211 ≈ ln e-.046t (take the ln of both sides) ln .211 ≈ -.046t ln e -1.556 ≈ -.046t (1) 33.8 ≈ t about 34 minutes to cool!

Domains of log equations Find the domain of # 19 on page 331 Domains are important because solving logarithmic equations sometimes produces extraneous roots.

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

Solve by decompressing log3(5x-1) = log3(x+7) Solve by decompressing

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

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 use base 5 on both sides

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

log5x + log(x+1)=2 Decompress and exponentiate

log5x + log(x+1)=2 10log5x +5x = 102 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 4=x is the only solution 2

One More! log2x + log2(x-7) = 3 Solve and check:

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

Page 331 # 38 Solve and ck. log(x-2) + log(x+5) =2log 3

Solution: log(x-2) + log(x+5) =2log 3 (x-2)(x+5)=9 x2 + 3x -19 = 0 (the neg. answer is extraneous)

Assignment Hw pg 331 18-25, 33-37 odds