View Full Version : solving logarithms
kdinser
Apr11-04, 07:15 AM
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
Apr11-04, 07:35 AM
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?
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
Apr11-04, 07:41 AM
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.
The inverse of g(x = 4) is 1/(7 + e4) though.
matt grime
Apr11-04, 08:16 AM
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
Apr11-04, 10:04 AM
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
Apr11-04, 12:01 PM
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+ ex= 4 or
x+ ex= 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
Apr11-04, 12:25 PM
Thanks guys, after thinking about Matt's answer, I'm pretty sure that's what the problem was getting at.
kim1214
Aug19-08, 09:57 AM
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+ ex= 4 or
x+ ex= 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
Aug22-08, 05:23 AM
If you will post your problem in a new thread I am sure a lot of people can help.
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