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  4) -ln(2)/3

  5) x = 9

  6) x = 2

  7) x = -6

Step-by-step explanation:

4) e^(-3x) = 2 . . . . given

  -3x = ln(2) . . . . . take the natural log

  x = -ln(2)/3 . . . . divide by the coefficient of x

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5) ln(3x -5) = ln(11) +ln(2)

  3x -5 = 11·2 . . . . . take antilogs

  3x = 27 . . . . . . . . add 5

  x = 9 . . . . . . . . . . divide by 3

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6) log4(x -1) +log4(x +2) = 1

  (x -1)(x +2) = 4 . . . . . take antilogs

  x^2 +x -2 = 4 . . . . . . eliminate parentheses

  x^2 +x +1/4 = 6 1/4 . . . add 2 1/4 to complete the square

  (x +0.5)^2 = 6.25 . . . . write as a square

  x +0.5 = ±2.5 . . . . . . . take the square root

  x = -0.5 ± 2.5 = {-3, +2} . . . . . . x = -3 is an extraneous solution

  x = 2

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7) 8^x = 4^(x -3)

  2^(3x) = 2^(2(x -3)) . . . . write using a common base of 2

  3x = 2(x -3) . . . . . . . . . . take logarithms base 2

  x = -6 . . . . . . . . . . . . . . .subtract 2x

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step-by-step explanation: