Finding the inverse of y=3+x+e^x, which includes both x and e^x

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Linty Fresh

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Sep 6, 2005
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OK, so I want to find the inverse of the function:

y=3+x+e^x
I started out by rewriting the equation as
x=3 + y + e^y, solve for y

Then:
ln(x)=ln(3) + ln(y) + y(ln(e))
ln(x)=ln(3) + ln(y) + y


And here's where I get stuck. I tried isolating "y":

ln(y) + y = ln(x)-ln(3)


But where do I go from here? I mean ln(x)-ln(3) ----> ln(x/3), but how does that help.?

Thanks!
 
Linty Fresh said:
OK, so I want to find the inverse of the function:

y=3+x+e^x
I started out by rewriting the equation as
x=3 + y + e^y, solve for y

Then:
ln(x)=ln(3) + ln(y) + y(ln(e)) NO
ln(x)=ln(3) + ln(y) + y


And here's where I get stuck. I tried isolating "y":

ln(y) + y = ln(x)-ln(3)


But where do I go from here? I mean ln(x)-ln(3) ----> ln(x/3), but how does that help.?

Thanks!
Click to expand...
Sorry but ln( A+ B + C) \(\displaystyle \neq\) ln (A) + ln(B) + ln(C).
It is true that ln(ABC) = ln (A)* ln(B)*ln(C)


EDIT: I meant to say that ln(ABC) = ln (A) + ln(B) + ln(C)
 
Last edited:
Linty Fresh said:
OK, so I want to find the inverse of the function:

y=3+x+e^x
I started out by rewriting the equation as
x=3 + y + e^y, solve for y

Then:
ln(x)=ln(3) + ln(y) + y(ln(e))
ln(x)=ln(3) + ln(y) + y

And here's where I get stuck. I tried isolating "y":

ln(y) + y = ln(x)-ln(3)

But where do I go from here? I mean ln(x)-ln(3) ----> ln(x/3), but how does that help.?
Click to expand...

None of your work, starting with the ln, is right. But more than that, what you want to do (almost certainly) can't be done.

Generally, when the variable appears both in an exponent, and elsewhere, you can't solve algebraically. The function has an inverse function because it is one-to-one, but its inverse can't be written in closed form.

Can you give any context? Why do you want to do this?
 
Thanks. So am I barking up the wrong tree by taking the natural log of both sides?

ln(x) = ln(3+y+e^y)
 
Dr.Peterson said:
None of your work, starting with the ln, is right. But more than that, what you want to do (almost certainly) can't be done.

Generally, when the variable appears both in an exponent, and elsewhere, you can't solve algebraically. The function has an inverse function because it is one-to-one, but its inverse can't be written in closed form.

Can you give any context? Why do you want to do this?
Click to expand...

Sorry, missed this before replying.

The entire problem states:

If g(x) = 3 + x + e^x, find Inverse g(4)

Maybe I was reading too much into the problem.
 
Linty Fresh said:
Sorry, missed this before replying.

The entire problem states:

If g(x) = 3 + x + e^x, find Inverse g(4)

Maybe I was reading too much into the problem.
Click to expand...

Ah! Sometimes, you can find a value of a function for a specific input, though you can't write a general expression for the function.

So if we want g-1(4) = x, what is that equivalent to in terms of g(x)? Can you, by trial and error, solve that? (Hint: only one value of x will give a nice, neat value for ex.)

This is a good example of the reason our guidelines ask you to state the entire problem, not just the part you think you need!
 
Linty Fresh said:
Sorry, missed this before replying.

The entire problem states:

If g(x) = 3 + x + e^x, find Inverse g(4)

Maybe I was reading too much into the problem.
Click to expand...
I take it you mean to find g-1(4). 1st x = g-1(y). That is, g-1(4) is an x value and 4 is a y value.

You are told that y=4 and want to find x.

Can you proceed from here? Hint: Just use g(x) = 3 + x + e^x
 
Yeah, got it. I have a knack for making math a lot harder than it has to be.

Thanks, everyone! :wink:
 
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