two precalc?

Question

1) Solve for y
log(3)y+3log(3) (y^2)=14

log(3) means log base 3

2) show that f(x) = (x-2)/(x+3) is invertible by finding its inverse.

(a) what is domain of f(x)? range of inverse?
(b) domain of inverse? range of f(x)?

Answer 1

log_3 (y) + 3 log_3 (y^2) = 14
log_3 (y) + log_3 (y^2)^3 = 14
log_3 (y) + log_3 (y^6) = 14
log_3 (y*y^6) = 14
log_3 (y^7) = 14
y^7 = 3^14
(y^7)^(1/7) = (3^14)^(1/7)
y = 3^[(14)(1/7)] = 3^2 = 9
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the notation needed for the second problem cannot be done in this text box. assume that f-inverse is g in the following.

f[g(x)] = x {by definition}
[g(x) - 2] / [g(x) + 3] = x
g(x) - 2 = x [g(x) + 3] {cross multiply}
g(x) - 2 = xg(x) + 3x
g(x) - xg(x) = 3x + 2 {subtract xg(x) and add 2}
[g(x)](1 - x) = 3x + 2 {factor out g(x)}
g(x) = (3x + 2) / (1 - x) {divide by (1 - x)}

a) x can be anything except -3
b) x can be anything except 1

Answer 2

1) Solve for y
log(3)y + 3log(3)(y²) = 14
log(3)y + log(3)[(y²)³] = 14
log(3)y + log(3)(y^6) = 14
log(3)[y*(y^6)] = 14
log(3)(y^7) = 14

Exponentiate.

y^7 = 3^14
y = 3² = 9
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2) show that f(x) = (x - 2)/(x + 3) is invertible by finding its inverse.

Solve for y to find the inverse function.

x = (y - 2)/(y + 3)
x(y + 3) = y - 2
xy + 3x = y - 2
xy - y = -3x - 2
y - xy = 3x + 2
(1 - x)y = 3x + 2
y = (3x + 2)/(1 - x)

The inverse function is:

f^(-1)(x) = (3x + 2)/(1 - x)

(a) what is domain of f(x)?

f(x) = (x - 2)/(x + 3)
Domain is (-∞, -3) U (-3, ∞)

Range of inverse?

f^(-1)(x) = (3x + 2)/(1 - x)
Range is (-∞, ∞)

(b) Domain of inverse?
Domain is (-∞, 1) U (1, ∞)

Range of f(x)?
Range is (-∞, ∞)

Answer 3

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