Analytical Solutions for Exponential and Logarithmic Equations

1. Unit 4Solving Exponential and Logarithmic Equations b. Solve polynomial, exponential, and logarithmic equations analytically, graphically, and using appropriate technology.

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. Your turn! • 24x = 32x-1 • 24x = (25)x-1 • 4x = 5x-5 • 5 = x 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 • log2x = log7 • Xlog2=log7 • x = • ≈ 2.807

6. 4x = 15

7. 102x-3+4 = 21 • -4 -4 • 102x-3 = 17 • log10102x-3 = log1017 • 2x-3 = log 17 • 2x = 3 + 1.23 • x = 4.23/2 • ≈ 2.115

8. 5x+2 + 3 = 25 • 5x+2 = 22 • log5x+2 = log22 • (x+2)log5 = log22 x+2 = (log22/log5) x+2 = 1.92 • x ≈ -.08

9. 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

10. 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

11. 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

12. log5(3x + 1) = 2 • 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

13. log5x + log(x+1)=2 • log (5x)(x+1) = 2 (product property) • log (5x2 + 5x) = 2 • 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 4=x is the only solution

14. 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

15. Assignment