Exponential and Log Equations   Equations that Involve Logs Step by Step Method     Step 1:  Contract to a single log.     Step 2: Get the log by itself.     Step 3:  Exponentiate both sides with the appropriate base.     Step 4:  Solve.     Step 5:  Check your solution for domain errors. Example:   Solve    log5 x + log5 (x + 2) = log5 (x + 6) log5 x + log5 (x + 2) - log5 (x + 6) = 0 log5 x (x + 2) - log5 (x + 6) = 0 log5 x (x + 2)/(x + 6) = 0 Already done. x(x + 2)/(x + 6) = 50 = 1 x(x + 2) = x + 6 x2 + 2x - x - 6 = 0 x2 + x - 6 = 0 (x - 2)(x + 3) = 0 x = 2 or x = -3 Note that -3 is not in the domain of the first log hence the only solution is x = 2. Exercises:  Solve log(x + 6) + 1 = 2log(3x - 2) 1/2 log(x + 3) + log2 = 1 Exponential Equations Step 1:  Isolate the exponential Step 2:  Take the appropriate log of both sides. Step 3:  Solve Example: Solve   4e-7x = 15  e-7x = 15/4 lne-7x = ln(15/4) -7x = ln(15/4) x = ln(15/4)/-7   Exercises:   Solve 1 + 2ex = 9 (10x - 4)/e2x - 4 = 0 (lnx)2  = ln(x2) 23x + 4(2-3x) = 5    Application All living beings have a certain amount of radioactive carbon C14 in their bodies.  When the being dies the C14 slowly decays with a half life of about 5600 years.  Suppose a skeleton is found in Tahoe that has 42% of the original C14.   When did the person die? Solution: We can use the exponential decay equation:         y = Cekt  After 5600 years there is          C/2  C14 left.  Substituting, we get:         C/2 = Cek(5600) Dividing by C,         1/2 = e5600k Take ln of both sides,         ln(.5) = 5600k so that         k = [ln(.5)]/5600 = -.000124 The equation becomes         y = Ce-.000124t  To find out when the person died, substitute          y = .42C  and solve for t:         .42C = Ce-.000124t Divide by C,         .42 = e-.000124t Take ln of both sides,         ln(.42) = -.000124t Divide by -.000124         t = [ln(.42)]/(-.000124) = 6995 The person died about 7,000 years ago.