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  f⁻¹(x) = log(x)/log(16)

  g⁻¹(x) = (2^x -1)/3

Step-by-step explanation:

In each case, you want to solve ...

  x = f(y)

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a. x = 4^(2y)

  log(x) = 2y·log(4) . . . . . . . . take logs

  y = log(x)/(2·log(4)) . . . . . . .divide by the coefficient of y

  f⁻¹(x) = log(x)/log(16) . . . . simplify (4^2 = 16)

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b. x = g(y)

  x = log(3y +1)/log(2) . . . . use the change of base relation

  x·log(2) = log(3y +1) . . . . multiply by log(2)

  2^x = 3y +1 . . . . . . . . . . . take antilogs

  2^x -1 = 3y . . . . . . . . . . . subtract 1

  y = (2^x -1)/3 . . . . . . . . . . divide by the coefficient of y

  g⁻¹(x) = (2^x -1)/3 . . . . . . . . note that "-1" is not part of the exponent of 2

look a the sides they are perpendicular