# Section 9.8 Exponential & Logarithmic Equations

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```					Section 9.8
Exponential & Logarithmic Equations
   Notation: log10x = log x and logex = ln x
   Solving Exponential Equations
   Solving Logarithmic Equations
Solving Exponential Equations
     Exponent Property of Equality:
    bx = by is equivalent to x = y
     Practice:
     3x+1 = 81 3x+1 = 34  x+1 = 4                       x=3
     53x-4 = 25 53x-4 = 52  3x-4 = 2                     x=2
1
2   x2 2 x
      2   x2 2 x
2   1
  x  2x  1  0
2
x  1
2
Solving Logarithmic Equations
   Logarithmic Property of Equality
logbx = logby is equivalent to x = y
   3x = 5     log 3x = log 5
x·log 3 = log 5 x = log 5 / log 3 x ≈ 1.464973521
Check: 31.465 ≈ 5.000145454
   6x-3 = 2x
log 6x-3 = log 2x
(x-3)·log 6 = x·log 2
x·log 6 - 3·log 6 = x·log 2
x·log 6 - x·log 2 = 3·log 6
x·(log 6 - log 2) = 3·log 6
x = 3·log 6 / (log 6 - log 2) ≈ 4.8928
Solving Logarithmic Equations

   log 5x = 3
103 = 5x 1000 = 5x 200 = x
   log (3x+2) = log (2x-3)
3x + 2 = 2x – 3  x = -5
log (-15+2) = log -13 which cannot exist
   log x + log (x-3) = 1 (1 = log 10)
log x(x-3) = log 10
x(x-3) = 10
x2 – 3x – 10 = 0
(x – 5)(x + 2) = 0  x = 5 and x = -2
discard x = -2 … only solution is x = 5
   A = A02-t/h
   Where A is the amount of radioactive material after t years,
A0 was the amount present at t = 0, and h is the material’s ½ life
   How old is a wooden statue when it contains only 1/3 of
its carbon-14 content (carbon-14 ½-life is 5700 years)
   ⅓A0 = A02-t/5700
1/3 = 2-t/5700
1 =3(2-t/5700) log 1 =log (3·2-t/5700)
0 = log 3 + log 2-t/5700
-log 3 = -(t/5700)·log 2
(5700·log 3)/log 2 = t    t ≈ 9,034.28 about 9000 years
What Next? Conic Sections
   Present Chapter 10

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