9514 1404 393
Answer:
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
