Inverse functions: find the inverse of h(x) = x/2x+1

[santaclaus]

I'm trying to find the inverses of these functions:

1. h(x)=x/2x+1
first i solved for y=x/2x+1 and then I changed x and y values. Eventually got h^-1(x)=x(2y+1)

2. h(x)= log (x-3)
do I take log x and log3?

3. q(x)=ln(x+3) - ln(x-5)
do I distribute here?

4. s(x)=3/2+logx
no idea

[Martingale]

I'm trying to find the inverses of these functions:

1. h(x)=x/2x+1
first i solved for y=x/2x+1 and then I changed x and y values. Eventually got h^-1(x)=x(2y+1)


2. h(x)= log (x-3)
do I take log x and log3?

3. q(x)=ln(x+3) - ln(x-5)
do I distribute here?

4. s(x)=3/2+logx
no idea

interchange and



solve for

and get



for logs



and

[stapel_eliz]

I'm trying to find the inverses of these functions:
2. h(x)= log (x-3)
do I take log x and log3?

To learn the basic process for finding inverses of functions, try here. To learn how to solve log equations, try here.

Once you have learned the necessary background material, the following should make sense.

Rename "h(x)" as "y".

Solve the log equation for "x=".

Swap the variables.

Rename the new "y" as "h^-1(x)".

3. q(x)=ln(x+3) - ln(x-5)
do I distribute here?

I don't know what you mean by "distributing"...?

To learn the rules for how to combine two log terms into one, try here.

Once you have learned the necessary background material:

Apply a log rule to combine the subtracted log terms into one log term.

Rename "q(x)" as "y".

Solve the resulting log equation for "x=".

Swap the variables.

Rename the new "y" as "q^-1".

4. s(x)=3/2+logx
no idea

Use the definition of logarithms to convert the fraction 3/2 into a logarithm. Then proceed as above.

If, after you have thoroughly studied all the lessons, you get stuck, please reply showing your work and reasoning so far. Thank you!