Steven, first of all, exchange for all *x*'s for *y*'s.

*y* = (2*e*^{x}- 3)/(7*e*^{x} + 9) => *x* = (2*e*^{y} - 3)/(7*e*^{y}+ 9)

The next step is to multiply both sides for 7*e*^{y }+ 9; it's supposed to give you

*x*(7*e*^{y}+ 9) = 2*e*^{y} - 3

Apply now the distributive property for the left side of the equation.

7*xe*^{y} + 9*x* = 2*e*^{y}* *- 3

Next, subtract 2*e*^{y}on both sides.

7*xe*^{y} + 9*x* - 2*e*^{y}= 2*e*^{y}- 3 - 2*e*^{y}

7*xe*^{y} + 9*x* - 2*e*^{y} = ^{-}3

Next, subtract 9*x* on both sides.

7*xe*^{y} + 9*x* - 2*e*^{y} - 9*x* = ^{-}3 - 9*x*

7*xe*^{y} - 2*e*^{y} = ^{-}9*x* - 3

Having the terms with *e*^{y}on one side of the equation, factor the left side using, precisely, *e*^{y} as the common factor.

*e*^{y}(7*x* - 2) = ^{-}9*x* - 3

Now divide each side by 7*x* - 2; the equation now must be solved for *e*^{y}.

*e*^{y} = (^{-}9*x* - 3)/(7*x* - 2)

It is preferable to factor the numerator using ^{-}1 as the common factor. Note: Brackets instead of parentheses where needed in order to distinguish the numerator from the denominator.

*e*^{y} = [^{-}1(3 - 9*x*)]/(7*x* - 2)

Now, apply the ^{-}1 to the denominator.

*e*^{y} = (3 - 9*x*)/(2 - 7*x*)

For the final step, you must get rid of the *e* in *e*^{y}. For this to happen, you must use the natural logarithm (*ln*), which is some sort of an opposite operation for *e*. Note: The brackets are also used for this step, this time to distinguish the content of the natural logarithm.

*y* = *ln*[(3 - 9*x*)/(2 - 7*x*)]

Seeing that this equation is finally solved for *y*, you might say that this is the right answer, but, to make this result a lot more classy, let's use the property of logarithms concerning division, that is, *log*(*a*/*b*) = *log*(*a*) - *log*(*b*). Applying this law to the result, this gives you

*y* = *ln*(3 - 9*x*) - *ln*(2 - 7*x*)

This result is the inverse of the original exercise.