Solving Logarithms: Finding the Inverse of g(x)=3+x+e^x

[kdinser]

Greetings all, I'm doing a refresh of calculus and physics in preparation for getting back to school this fall after a 5 year layoff. Most stuff is coming back pretty quickly, but I'm stuck on this one problem. I'm sure I'm missing something small, but I just haven't been able to find any example problems that match this one.

g(x)=3+x+e^x
g(x)=y

I need to find the inverse of this function, which means solving for x. If I take the ln e^x to get x, I'm stuck on the other side with ln (y-3-x)=x I'm not sure where to go from here, any help would be greatly appreciated.

 

[HallsofIvy]

Since g has x both "inside" and "outside" the exponential, its inverse is not any "elementary" function. Exactly what does the problem ask you to do?

 

[Zurtex]

If your finding the root of the equation then you want to find x for:

$$3 + x + e^x = 0$$

And I must say I'm a little stumped on how you would work out an exact answer.

 

[kdinser]

if g(x)=3+x+e^x Find g^-1(4)

I'm reading this as "find the inverse g(x) function and then solve it when x=4" The answer in the back of the book shows that it should be 0.

 

[Chen]

The inverse of g(x = 4) is 1/(7 + e⁴) though.

 

[matt grime]

that isn't the inverse, that's the reciprocal.

by inspection g(0) = 3+0+e^0 = 4,

(technical note, g is monotone increasing so inverse is well defined)

hence g^{-1}(4) = 0

and similiarly g^{-1}(4+e) = 1

 

[philosophking]

Since it wasn't immediately apparent to me what matt was doing, here's a general point:

For some function f(x),

If:
f(b)=a

Then:
f^{-1} (a) = b

 

[HallsofIvy]

Yep. The problem did NOT ask that you actually find g^-1(x),
only that you find g^-1(4).

That is exactly the same as solving the equation 3+ x+ e^x= 4 or
x+ e^x= 1. There is still no general way of solving an equation like that, but you might be as smart as Matt Grime and recognize that if x= 0, e⁰= 1 so
0+ e⁰= 1. g^-1(4)= 0.

 

[kdinser]

Thanks guys, after thinking about Matt's answer, I'm pretty sure that's what the problem was getting at.

 

[kim1214]

Yep. The problem did NOT ask that you actually find g^-1(x),
only that you find g^-1(4).

That is exactly the same as solving the equation 3+ x+ e^x= 4 or
x+ e^x= 1. There is still no general way of solving an equation like that, but you might be as smart as Matt Grime and recognize that if x= 0, e⁰= 1 so
0+ e⁰= 1. g^-1(4)= 0.

need help solving a problem can u help

 

[HallsofIvy]

If you will post your problem in a new thread I am sure a lot of people can help.