College Algebra question(HElp, seems like no one wnats to answer these anymore)?

2 log(x+1) = 4 - f

x = exp( 2 - f/2) - 1

10^((4-x)/2) - 1 seems to be the popular answer.

ok, to make it simpler, lets use just y instead of having to type f(x) every time.

y=4-2log(x+1)

to get the inverse, we must find the equation in the form of x=, and this will be our inverse.

2log(x+1)+y=4

2log(x+1)=4-y

log(x+1)=(4-y)/2=2-y/2

10^log(x+1)=10^(2-y/2)

x+1=10^(2-y/2)

x=10^(2-y/2)-1

or, when the exponent is as one fraction:

x=[10^{(4-y)/2}]-1

as this is true for the real equation, switching the x and y will make it true for the inverse equation:

y=[10^{(4-x)/2}]-1

and i guess thats bout it.

You have:

y = 4 - 2log(x + 1)

Replace x with y:

x = 4 - 2log(y + 1)

so:

2log(y + 1) = 4 - x

log (y+1)^2 = 4 - x

(y + 1)^2 = e^(4-x)

y + 1 = sqrt(e^(4-x))

So:

y = sqrt(e^(4-x)) - 1

is the inverse

y = 4 - 2log(x + 1)

2log(x + 1) = 4 - y

log(x + 1) = (4 - y)/2

10^log(x + 1) = 10^[(4 - y)/2]

x + 1 = 10^[(4 - y)/2]

x = 10^[(4 - y)/2] - 1

y = 10^[(4 - x)/2] - 1

y = 4 - 2log(x+1)

2log(x+1) = 4-y

log(x+1) = (4 - y)/2

x + 1 = 10^((4-y)/2)

x = 10^((4-y)/2) - 1

therefore,

f_inverse(x) = 10^((4-x)/2) - 1

Ans

The reason no one seems to want to answer these questions any more is twofold: most people on Y!A don't appear to be very smart and many of us believe people should do their own homework!