1. 7.6A Solving Exponential and Logarithmic Equations Algebra II

2. 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 b x = b y, then x=y Exponential Equations

3. Ex. 1)Solve by equating exponents 4 3x = 8 x+1 (2 2 ) 3x = (2 3 ) x+1 rewrite w/ same base 2 6x = 2 3x+3 6x = 3x+3 x = 1 Check → 4 3*1 = 8 1+1 64 = 64

4. Ex. 2) Solve by equating exponents 2 4x = 32 x-1 2 4x = (2 5 ) x-1 4x = 5x-5 5 = x Be sure to check your answer!!!

5. Ex. 3 When you can’t rewrite using the same base, you can solve by taking a log of both sides 2 x = 7 log 2 2 x = log 2 7 x = log 2 7 x = ≈ 2.807

6. Ex. 4) Solve by equating exponents log 4 4 x = log 4 15 x = log 4 15 = log15/log4 ≈ 1.953

7. Ex. 5 5 x+2 + 3 = 25 5 x+2 = 22 log 5 5 x+2 = log 5 22 x+2 = log 5 22 x = (log 5 22) – 2 = (log22/log5) – 2 ≈ -.079

8. Ex.6 10 2x-3 +4 = 21 -4 -4 10 2x-3 = 17 log 10 10 2x-3 = log 10 17 2x-3 = log 17 2x = 3 + log17 x = ½(3 + log17) ≈ 2.115

9. Assignment

10. 7.6B Solving Log Equations To solve use the property for logs w/ the same base: + #’s b,x,y & b≠1 If log b x = log b y, then x = y

11. Ex. 1 log 3 (5x-1) = log 3 (x+7) 5x – 1 = x + 7 5x = x + 8 4x = 8 x = 2 and check log 3 (5*2-1) = log 3 (2+7) log 3 9 = log 3 9

12. When faced with log/logs on one side of equation, then exponentiate each side. b>0 & b≠1 if x = y, then b x = b y

13. Ex. 2) log 5 (3x + 1) = 2 5 log 5 (3x+1) = 5 2 3x+1 = 25 x = 8 and check Because the domain of log functions doesn’t include all reals, you should check for extraneous solutions

14. Ex. 3 log 2 x + log 2 (x-7) = 3 log 2 x(x-7) = 3 log 2 (x 2 - 7x) = 3 2 log 2 x² -7x = x 2 – 7x = 8 x 2 – 7x – 8 = 0 (x-8)(x+1)=0 x=8 x= -1

15. Ex. 4 log5x + log(x-1)=2 log (5x)(x-1) = 2 (product property) log (5x 2 – 5x) = 2 10 log5x -5x = 10 2 5x 2 - 5x = 100 x 2 – 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

16. EX. 5 Newton’s Law of Cooling The temperature T of a cooling substance @ time t (in minutes) is: T = (T 0 – T R ) e -rt + T R T 0 = initial temperature T R = room temperature r = constant cooling rate of the substance

17. 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°?

18. T 0 = 212, T R = 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?

19. 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.