8.6 Solving Exponential and Logarithmic Equations 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?

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

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

24x = 32x-1 24x = (25)x-1 4x = 5x-5 5 = x Your turn! 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 = ≈ 2.807 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 -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 = (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 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) .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 Cooling cont. 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!

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

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

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

3x+1 = 25 x = 8 and check 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