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

the multiples of 15 are 15, 30,45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195,   210, 225, 240, 255, 270, 285, 300, 315, 330, 345, 360, 375,   390, 405, 420, 435, 450, 465, 480, 495, 510, 525, 540, 555, 570, 585, 600, 615, 630, 645, 660, 675, 690, 705, 720, and 735.

the factors of 15 are 1, 3, 5, and 15.

hope that , fam!

-une, your fren : )