Find the inverse of each function.

a. f (x) = 3x b. f (x) = (1/2) ^x

c. g (x) = ln (x - 7) d. h (x) = log3 (x + 2) / log3 (5)

e. f (x) = 3 (1.8) 0.2x + 3 f. g (x) = log2 (3√x - 4)

g. h (x) = 5x / 5x + 1

h. f (x) = 2-x+1

a. f (x) ^ (-1) = x/3

b. f (x) ^ (-1) = log_10⁡x/log_10⁡〖1/2〗

c. g (x) ^ (-1) = e^x+7

d. f (x) ^ (-1) = 5^x-2

e. f (x) ^ (-1) = (x-3) / 1.08

f. f (x) ^ (-1) = ((2^x-4) / 3) ^2

g. h (x) ^ (-1) = x / (5-5x)

h. f (x) ^ (-1) = x+3

Step-by-step explanation:

To understand how to find the inverse function you must solve the equation for x and then replace x by the definition of the inverse f (x) - 1. Then the explanation of each exercise goes as follows:

a. f (x) = 3x

f (x) / 3=x

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

b. f (x) = (1/2) ^ (x)

Considering that log_b⁡ (a) = c and log_b⁡ (a) = log_c⁡ (a) / log_c⁡ (b)

log_ (1/2) f (x) = log_ (1/2) ⁡ ((1/2) ^x)

log_ (1/2) f (x) = x

f (x) ^ (-1) = log_ (1/2) (x) = log_10⁡ (x) / log_10⁡ (1/2)

c. g (x) = ln⁡ (x-7)

Considering that ln⁡ (e^x) = x

e^g (x) = e^ln⁡ (x-7)

e^g (x) = (x-7)

x=e^g (x) + 7

g (x) ^ (-1) = e^x+7

d. h (x) = log_3⁡ (x+2) / log_3⁡ (5)

log_3⁡ (x+2) / log_3⁡ (5) = log_5⁡ (x+2)

h (x) = log_5⁡ (x+2)

5^h (x) = x+2

h (x) ^ (-1) = 5^x-2

e. f (x) = 3*1.8*0.2*x+3

f (x) = 3*1.8*0.2*x+3

f (x) ^ (-1) = (x-3) / 1.08

f. f (x) = log_2⁡ (3√x-4)

2^f (x) = 2^log_2⁡ (3√x-4)

2^f (x) - 4=3√x

((2^f (x) - 4) / 3) ^2=√x^2

f (x) ^ (-1) = ((2^x-4) / 3) ^2

g. h (x) = 5x / (5x+1)

(5x+1) h (x) = 5x

h (x) = 5x (1-h (x))

x=h (x) / 5 (1-h (x))

h (x) ^ (-1) = x / (5 (1-x))

h. f (x) = 2-x+1=3-x

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