function inverses with exponential, why is $x = 0$?
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I'm trying to learn how to use function inverses to solve equations. In an exercise I know that the solution is $0$. I am unable to reach zero myself.
Solve for $x$ in the equation $g(x) = 1$, given $g(x) = e^{-2x}$.
From my book I know that the inverse of $e^x$ is $log_e(x)$.
The solution in my textbook reads:
$$g^{-1}(x)=-frac{1}{2}log(x), x=0.$$
I'm unable to follow or understand the steps to arrive at $x = 0$?
functions inverse-function
edited Nov 20 at 20:45
Robert Howard
1,9141822
asked Aug 18 at 20:40
Doug Fir
1696
add a comment |
up vote
1
down vote
favorite
I'm trying to learn how to use function inverses to solve equations. In an exercise I know that the solution is $0$. I am unable to reach zero myself.
Solve for $x$ in the equation $g(x) = 1$, given $g(x) = e^{-2x}$.
From my book I know that the inverse of $e^x$ is $log_e(x)$.
The solution in my textbook reads:
$$g^{-1}(x)=-frac{1}{2}log(x), x=0.$$
I'm unable to follow or understand the steps to arrive at $x = 0$?
functions inverse-function
edited Nov 20 at 20:45
Robert Howard
1,9141822
asked Aug 18 at 20:40
Doug Fir
1696
Surely, you have copied that wrong! Or it is a misprint in your text book. It should be "$xne 0$" because log(x) is not defined for x= 0.
– user247327
Aug 18 at 20:53
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm trying to learn how to use function inverses to solve equations. In an exercise I know that the solution is $0$. I am unable to reach zero myself.
Solve for $x$ in the equation $g(x) = 1$, given $g(x) = e^{-2x}$.
From my book I know that the inverse of $e^x$ is $log_e(x)$.
The solution in my textbook reads:
$$g^{-1}(x)=-frac{1}{2}log(x), x=0.$$
I'm unable to follow or understand the steps to arrive at $x = 0$?
functions inverse-function
edited Nov 20 at 20:45
Robert Howard
1,9141822
asked Aug 18 at 20:40
Doug Fir
1696
I'm trying to learn how to use function inverses to solve equations. In an exercise I know that the solution is $0$. I am unable to reach zero myself.
Solve for $x$ in the equation $g(x) = 1$, given $g(x) = e^{-2x}$.
From my book I know that the inverse of $e^x$ is $log_e(x)$.
The solution in my textbook reads:
$$g^{-1}(x)=-frac{1}{2}log(x), x=0.$$
I'm unable to follow or understand the steps to arrive at $x = 0$?
functions inverse-function
functions inverse-function
edited Nov 20 at 20:45
Robert Howard
1,9141822
asked Aug 18 at 20:40
Doug Fir
1696
edited Nov 20 at 20:45
Robert Howard
1,9141822
asked Aug 18 at 20:40
Doug Fir
1696
edited Nov 20 at 20:45
Robert Howard
1,9141822
edited Nov 20 at 20:45
Robert Howard
1,9141822
edited Nov 20 at 20:45
Robert Howard
1,9141822
1,9141822
asked Aug 18 at 20:40
Doug Fir
1696
asked Aug 18 at 20:40
Doug Fir
1696
asked Aug 18 at 20:40
Doug Fir
1696
1696
Surely, you have copied that wrong! Or it is a misprint in your text book. It should be "$xne 0$" because log(x) is not defined for x= 0.
– user247327
Aug 18 at 20:53
add a comment |
Surely, you have copied that wrong! Or it is a misprint in your text book. It should be "$xne 0$" because log(x) is not defined for x= 0.
– user247327
Aug 18 at 20:53
Surely, you have copied that wrong! Or it is a misprint in your text book. It should be "$xne 0$" because log(x) is not defined for x= 0.
– user247327
Aug 18 at 20:53
Surely, you have copied that wrong! Or it is a misprint in your text book. It should be "$xne 0$" because log(x) is not defined for x= 0.
– user247327
Aug 18 at 20:53
add a comment |
1 Answer
1
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oldest
votes
up vote
2
down vote
accepted
$$y=e^{-2x}implies ln y=ln e^{-2x}=-2ximplies x=dfrac{ln y}{-2}=dfrac{-ln y}{2}.$$
- In other words, $g(x)=e^{-2x}implies g^{-1}(x)=dfrac{-ln x}{2}.$
- In the first step we have used that the inverse function of $e^x$ is $ln x=log_e x.$
- In the second step we just have isolated $x.$
Now, if $g(x)=1$ we get
$$x=g^{-1}(1)=dfrac{-ln 1}{2}=0.$$
answered Aug 18 at 20:44
mfl
26k12141
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$$y=e^{-2x}implies ln y=ln e^{-2x}=-2ximplies x=dfrac{ln y}{-2}=dfrac{-ln y}{2}.$$
- In other words, $g(x)=e^{-2x}implies g^{-1}(x)=dfrac{-ln x}{2}.$
- In the first step we have used that the inverse function of $e^x$ is $ln x=log_e x.$
- In the second step we just have isolated $x.$
Now, if $g(x)=1$ we get
$$x=g^{-1}(1)=dfrac{-ln 1}{2}=0.$$
answered Aug 18 at 20:44
mfl
26k12141
add a comment |
up vote
2
down vote
accepted
$$y=e^{-2x}implies ln y=ln e^{-2x}=-2ximplies x=dfrac{ln y}{-2}=dfrac{-ln y}{2}.$$
- In other words, $g(x)=e^{-2x}implies g^{-1}(x)=dfrac{-ln x}{2}.$
- In the first step we have used that the inverse function of $e^x$ is $ln x=log_e x.$
- In the second step we just have isolated $x.$
Now, if $g(x)=1$ we get
$$x=g^{-1}(1)=dfrac{-ln 1}{2}=0.$$
answered Aug 18 at 20:44
mfl
26k12141
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$$y=e^{-2x}implies ln y=ln e^{-2x}=-2ximplies x=dfrac{ln y}{-2}=dfrac{-ln y}{2}.$$
- In other words, $g(x)=e^{-2x}implies g^{-1}(x)=dfrac{-ln x}{2}.$
- In the first step we have used that the inverse function of $e^x$ is $ln x=log_e x.$
- In the second step we just have isolated $x.$
Now, if $g(x)=1$ we get
$$x=g^{-1}(1)=dfrac{-ln 1}{2}=0.$$
answered Aug 18 at 20:44
mfl
26k12141
$$y=e^{-2x}implies ln y=ln e^{-2x}=-2ximplies x=dfrac{ln y}{-2}=dfrac{-ln y}{2}.$$
- In other words, $g(x)=e^{-2x}implies g^{-1}(x)=dfrac{-ln x}{2}.$
- In the first step we have used that the inverse function of $e^x$ is $ln x=log_e x.$
- In the second step we just have isolated $x.$
Now, if $g(x)=1$ we get
$$x=g^{-1}(1)=dfrac{-ln 1}{2}=0.$$
answered Aug 18 at 20:44
mfl
26k12141
edited Aug 18 at 20:50
answered Aug 18 at 20:44
mfl
26k12141
answered Aug 18 at 20:44
mfl
26k12141
answered Aug 18 at 20:44
mfl
26k12141
26k12141
add a comment |
add a comment |
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Surely, you have copied that wrong! Or it is a misprint in your text book. It should be "$xne 0$" because log(x) is not defined for x= 0.
– user247327
Aug 18 at 20:53