A quick question about logs
$begingroup$
$$ln(e^x)=x$$
Is this right?
If yes, then please tell me how.
I know
$e^{ln x}$ $=x$.
calculus logarithms
edited Jan 22 at 22:30
psmears
70949
asked Jan 22 at 12:46
AashishAashish
828
$endgroup$
|
show 3 more comments
$begingroup$
$$ln(e^x)=x$$
Is this right?
If yes, then please tell me how.
I know
$e^{ln x}$ $=x$.
calculus logarithms
edited Jan 22 at 22:30
psmears
70949
asked Jan 22 at 12:46
AashishAashish
828
$endgroup$
6
$begingroup$
It's true, and the simple explanation is that ln and $e^x$ are each other's inverse functions. And there's no range problem because $e^x > 0$. Only in calculating the latter ($e^{ln{x}}$) you have to take $x>0$.
$endgroup$
– Matti P.
Jan 22 at 12:49
3
$begingroup$
Let's say $ln(e^x)=y$. Then $e^{ln(e^x)}=e^x=e^y$. Here I used your second line. Since the exponential function is injective it follows that $x=y$.
$endgroup$
– maxmilgram
Jan 22 at 12:55
$begingroup$
The history is worth considering: Natural log predated Euler's $e^x$
$endgroup$
– TurlocTheRed
Jan 22 at 22:24
$begingroup$
Are you assuming that $x$ is a real number?
$endgroup$
– Dan
Jan 22 at 23:43
$begingroup$
It is true iff $Im xin(-pi,pi]$. Assuming the branch of $ln$ is principal.
$endgroup$
– Kemono Chen
Jan 23 at 2:13
|
show 3 more comments
$begingroup$
$$ln(e^x)=x$$
Is this right?
If yes, then please tell me how.
I know
$e^{ln x}$ $=x$.
calculus logarithms
edited Jan 22 at 22:30
psmears
70949
asked Jan 22 at 12:46
AashishAashish
828
$endgroup$
$$ln(e^x)=x$$
Is this right?
If yes, then please tell me how.
I know
$e^{ln x}$ $=x$.
calculus logarithms
calculus logarithms
edited Jan 22 at 22:30
psmears
70949
asked Jan 22 at 12:46
AashishAashish
828
edited Jan 22 at 22:30
psmears
70949
asked Jan 22 at 12:46
AashishAashish
828
edited Jan 22 at 22:30
psmears
70949
edited Jan 22 at 22:30
psmears
70949
edited Jan 22 at 22:30
psmears
70949
70949
asked Jan 22 at 12:46
AashishAashish
828
asked Jan 22 at 12:46
AashishAashish
828
asked Jan 22 at 12:46
AashishAashish
828
828
6
$begingroup$
It's true, and the simple explanation is that ln and $e^x$ are each other's inverse functions. And there's no range problem because $e^x > 0$. Only in calculating the latter ($e^{ln{x}}$) you have to take $x>0$.
$endgroup$
– Matti P.
Jan 22 at 12:49
3
$begingroup$
Let's say $ln(e^x)=y$. Then $e^{ln(e^x)}=e^x=e^y$. Here I used your second line. Since the exponential function is injective it follows that $x=y$.
$endgroup$
– maxmilgram
Jan 22 at 12:55
$begingroup$
The history is worth considering: Natural log predated Euler's $e^x$
$endgroup$
– TurlocTheRed
Jan 22 at 22:24
$begingroup$
Are you assuming that $x$ is a real number?
$endgroup$
– Dan
Jan 22 at 23:43
$begingroup$
It is true iff $Im xin(-pi,pi]$. Assuming the branch of $ln$ is principal.
$endgroup$
– Kemono Chen
Jan 23 at 2:13
|
show 3 more comments
6
$begingroup$
It's true, and the simple explanation is that ln and $e^x$ are each other's inverse functions. And there's no range problem because $e^x > 0$. Only in calculating the latter ($e^{ln{x}}$) you have to take $x>0$.
$endgroup$
– Matti P.
Jan 22 at 12:49
3
$begingroup$
Let's say $ln(e^x)=y$. Then $e^{ln(e^x)}=e^x=e^y$. Here I used your second line. Since the exponential function is injective it follows that $x=y$.
$endgroup$
– maxmilgram
Jan 22 at 12:55
$begingroup$
The history is worth considering: Natural log predated Euler's $e^x$
$endgroup$
– TurlocTheRed
Jan 22 at 22:24
$begingroup$
Are you assuming that $x$ is a real number?
$endgroup$
– Dan
Jan 22 at 23:43
$begingroup$
It is true iff $Im xin(-pi,pi]$. Assuming the branch of $ln$ is principal.
$endgroup$
– Kemono Chen
Jan 23 at 2:13
6
6
$begingroup$
It's true, and the simple explanation is that ln and $e^x$ are each other's inverse functions. And there's no range problem because $e^x > 0$. Only in calculating the latter ($e^{ln{x}}$) you have to take $x>0$.
$endgroup$
– Matti P.
Jan 22 at 12:49
$begingroup$
It's true, and the simple explanation is that ln and $e^x$ are each other's inverse functions. And there's no range problem because $e^x > 0$. Only in calculating the latter ($e^{ln{x}}$) you have to take $x>0$.
$endgroup$
– Matti P.
Jan 22 at 12:49
3
3
$begingroup$
Let's say $ln(e^x)=y$. Then $e^{ln(e^x)}=e^x=e^y$. Here I used your second line. Since the exponential function is injective it follows that $x=y$.
$endgroup$
– maxmilgram
Jan 22 at 12:55
$begingroup$
Let's say $ln(e^x)=y$. Then $e^{ln(e^x)}=e^x=e^y$. Here I used your second line. Since the exponential function is injective it follows that $x=y$.
$endgroup$
– maxmilgram
Jan 22 at 12:55
$begingroup$
The history is worth considering: Natural log predated Euler's $e^x$
$endgroup$
– TurlocTheRed
Jan 22 at 22:24
$begingroup$
The history is worth considering: Natural log predated Euler's $e^x$
$endgroup$
– TurlocTheRed
Jan 22 at 22:24
$begingroup$
Are you assuming that $x$ is a real number?
$endgroup$
– Dan
Jan 22 at 23:43
$begingroup$
Are you assuming that $x$ is a real number?
$endgroup$
– Dan
Jan 22 at 23:43
$begingroup$
It is true iff $Im xin(-pi,pi]$. Assuming the branch of $ln$ is principal.
$endgroup$
– Kemono Chen
Jan 23 at 2:13
$begingroup$
It is true iff $Im xin(-pi,pi]$. Assuming the branch of $ln$ is principal.
$endgroup$
– Kemono Chen
Jan 23 at 2:13
|
show 3 more comments
5 Answers
5
active
oldest
votes
$begingroup$
Well:
$$ln(e^{ln(x)})=ln(x)ln(e)=ln(x)$$
By the logarithmic power rule. What you seek comes from this.
answered Jan 22 at 13:00
Rhys HughesRhys Hughes
6,6141530
$endgroup$
1
$begingroup$
Isn't the power rule a stronger property than what is to prove ?
$endgroup$
– Yves Daoust
Jan 23 at 9:22
$begingroup$
Perhaps. But it does it nicely and power rule is one of the first things learnt about logs.
$endgroup$
– Rhys Hughes
Jan 23 at 10:59
add a comment |
$begingroup$
Let $f: mathbb R to (0, infty)$ be defined by $f(x)=e^x$. Then $f$ is bijective, hence $f^{-1}: (0, infty) to mathbb R$ exists.
By definition(!): $ ln x :=f^{-1}(x)$ for $x>0.$
Hence $e^{ln x}=f(f^{-1}(x))=x$ for all $x>0$ and $ln(e^x)=f^{-1}(f(x))=x$ for all $x in mathbb R.$
answered Jan 22 at 12:50
FredFred
46.5k1848
$endgroup$
add a comment |
$begingroup$
For real numbers, it's true. But
$ln e^{2pi i}=ln 1 = 0$
This is due to the fact that $f: x rightarrow e^x$ is injective for real $x$, but not for complex $x$.
answered Jan 22 at 22:29
AcccumulationAcccumulation
6,9642618
$endgroup$
add a comment |
$begingroup$
Yes, this is correct. It is because the logarithm function is defined in that manner. A logarithm is essentially asking the question To what power $x$ do I raise some number $a$ to get $b$?
In general, the statements $a^x=b$ and $log_ab=x$ are equivalent.
In this case, since the base of the logarithm is $e$, you can ask yourself the same question: To what power do I raise $e$ to get $e^x$? Clearly the answer is $x$.
I recommend you read up on logarithms if you are having difficulty in this basic concept: https://en.m.wikipedia.org/wiki/Logarithm
answered Jan 22 at 12:54
Naman KumarNaman Kumar
19613
$endgroup$
add a comment |
$begingroup$
With $x=ln(y)$, $$e^{ln(y)}=yimplies ln(e^{ln(y)})=ln(y)implies ln(e^x)=x.$$
There is no need to refer to inversion nor any other property than the given.
answered Jan 22 at 12:58
Yves DaoustYves Daoust
127k673226
$endgroup$
1
$begingroup$
Haven't you used the following property without stating it? That is: for all x there exists y such x = ln(y); i.e. the logarithm function is onto the set in question (we can probably assume that set is the reals, although it hasn't been stated).
$endgroup$
– Don Hatch
Jan 22 at 22:25
$begingroup$
@DonHatch indeed, but do recall a function is invertible iff it is bijective. In order to talk of the logarithm as an inverse we should probably have valided the exponential is bijective on the reals. Over the complex numbers we lose injectivity, but preserve surjectivity, so we can use the concept of covering spaces or the like to define the "inverse" of a non-injective function.
$endgroup$
– Brevan Ellefsen
Jan 23 at 2:49
$begingroup$
@DonHatch: no, I cheated by saying "take $x=ln y$", so the proof is valid for all such pairs (if any), but only those. Anyway, if we add that the range of the logarithm is $mathbb R$, the proof holds for all real $x$. The range condition is sufficient, invertibility is not required.
$endgroup$
– Yves Daoust
Jan 23 at 9:13
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Well:
$$ln(e^{ln(x)})=ln(x)ln(e)=ln(x)$$
By the logarithmic power rule. What you seek comes from this.
answered Jan 22 at 13:00
Rhys HughesRhys Hughes
6,6141530
$endgroup$
1
$begingroup$
Isn't the power rule a stronger property than what is to prove ?
$endgroup$
– Yves Daoust
Jan 23 at 9:22
$begingroup$
Perhaps. But it does it nicely and power rule is one of the first things learnt about logs.
$endgroup$
– Rhys Hughes
Jan 23 at 10:59
add a comment |
$begingroup$
Well:
$$ln(e^{ln(x)})=ln(x)ln(e)=ln(x)$$
By the logarithmic power rule. What you seek comes from this.
answered Jan 22 at 13:00
Rhys HughesRhys Hughes
6,6141530
$endgroup$
1
$begingroup$
Isn't the power rule a stronger property than what is to prove ?
$endgroup$
– Yves Daoust
Jan 23 at 9:22
$begingroup$
Perhaps. But it does it nicely and power rule is one of the first things learnt about logs.
$endgroup$
– Rhys Hughes
Jan 23 at 10:59
add a comment |
$begingroup$
Well:
$$ln(e^{ln(x)})=ln(x)ln(e)=ln(x)$$
By the logarithmic power rule. What you seek comes from this.
answered Jan 22 at 13:00
Rhys HughesRhys Hughes
6,6141530
$endgroup$
Well:
$$ln(e^{ln(x)})=ln(x)ln(e)=ln(x)$$
By the logarithmic power rule. What you seek comes from this.
answered Jan 22 at 13:00
Rhys HughesRhys Hughes
6,6141530
answered Jan 22 at 13:00
Rhys HughesRhys Hughes
6,6141530
answered Jan 22 at 13:00
Rhys HughesRhys Hughes
6,6141530
answered Jan 22 at 13:00
Rhys HughesRhys Hughes
6,6141530
6,6141530
1
$begingroup$
Isn't the power rule a stronger property than what is to prove ?
$endgroup$
– Yves Daoust
Jan 23 at 9:22
$begingroup$
Perhaps. But it does it nicely and power rule is one of the first things learnt about logs.
$endgroup$
– Rhys Hughes
Jan 23 at 10:59
add a comment |
1
$begingroup$
Isn't the power rule a stronger property than what is to prove ?
$endgroup$
– Yves Daoust
Jan 23 at 9:22
$begingroup$
Perhaps. But it does it nicely and power rule is one of the first things learnt about logs.
$endgroup$
– Rhys Hughes
Jan 23 at 10:59
1
1
$begingroup$
Isn't the power rule a stronger property than what is to prove ?
$endgroup$
– Yves Daoust
Jan 23 at 9:22
$begingroup$
Isn't the power rule a stronger property than what is to prove ?
$endgroup$
– Yves Daoust
Jan 23 at 9:22
$begingroup$
Perhaps. But it does it nicely and power rule is one of the first things learnt about logs.
$endgroup$
– Rhys Hughes
Jan 23 at 10:59
$begingroup$
Perhaps. But it does it nicely and power rule is one of the first things learnt about logs.
$endgroup$
– Rhys Hughes
Jan 23 at 10:59
add a comment |
$begingroup$
Let $f: mathbb R to (0, infty)$ be defined by $f(x)=e^x$. Then $f$ is bijective, hence $f^{-1}: (0, infty) to mathbb R$ exists.
By definition(!): $ ln x :=f^{-1}(x)$ for $x>0.$
Hence $e^{ln x}=f(f^{-1}(x))=x$ for all $x>0$ and $ln(e^x)=f^{-1}(f(x))=x$ for all $x in mathbb R.$
answered Jan 22 at 12:50
FredFred
46.5k1848
$endgroup$
add a comment |
$begingroup$
Let $f: mathbb R to (0, infty)$ be defined by $f(x)=e^x$. Then $f$ is bijective, hence $f^{-1}: (0, infty) to mathbb R$ exists.
By definition(!): $ ln x :=f^{-1}(x)$ for $x>0.$
Hence $e^{ln x}=f(f^{-1}(x))=x$ for all $x>0$ and $ln(e^x)=f^{-1}(f(x))=x$ for all $x in mathbb R.$
answered Jan 22 at 12:50
FredFred
46.5k1848
$endgroup$
add a comment |
$begingroup$
Let $f: mathbb R to (0, infty)$ be defined by $f(x)=e^x$. Then $f$ is bijective, hence $f^{-1}: (0, infty) to mathbb R$ exists.
By definition(!): $ ln x :=f^{-1}(x)$ for $x>0.$
Hence $e^{ln x}=f(f^{-1}(x))=x$ for all $x>0$ and $ln(e^x)=f^{-1}(f(x))=x$ for all $x in mathbb R.$
answered Jan 22 at 12:50
FredFred
46.5k1848
$endgroup$
Let $f: mathbb R to (0, infty)$ be defined by $f(x)=e^x$. Then $f$ is bijective, hence $f^{-1}: (0, infty) to mathbb R$ exists.
By definition(!): $ ln x :=f^{-1}(x)$ for $x>0.$
Hence $e^{ln x}=f(f^{-1}(x))=x$ for all $x>0$ and $ln(e^x)=f^{-1}(f(x))=x$ for all $x in mathbb R.$
answered Jan 22 at 12:50
FredFred
46.5k1848
edited Jan 22 at 12:59
answered Jan 22 at 12:50
FredFred
46.5k1848
answered Jan 22 at 12:50
FredFred
46.5k1848
answered Jan 22 at 12:50
FredFred
46.5k1848
46.5k1848
add a comment |
add a comment |
$begingroup$
For real numbers, it's true. But
$ln e^{2pi i}=ln 1 = 0$
This is due to the fact that $f: x rightarrow e^x$ is injective for real $x$, but not for complex $x$.
answered Jan 22 at 22:29
AcccumulationAcccumulation
6,9642618
$endgroup$
add a comment |
$begingroup$
For real numbers, it's true. But
$ln e^{2pi i}=ln 1 = 0$
This is due to the fact that $f: x rightarrow e^x$ is injective for real $x$, but not for complex $x$.
answered Jan 22 at 22:29
AcccumulationAcccumulation
6,9642618
$endgroup$
add a comment |
$begingroup$
For real numbers, it's true. But
$ln e^{2pi i}=ln 1 = 0$
This is due to the fact that $f: x rightarrow e^x$ is injective for real $x$, but not for complex $x$.
answered Jan 22 at 22:29
AcccumulationAcccumulation
6,9642618
$endgroup$
For real numbers, it's true. But
$ln e^{2pi i}=ln 1 = 0$
This is due to the fact that $f: x rightarrow e^x$ is injective for real $x$, but not for complex $x$.
answered Jan 22 at 22:29
AcccumulationAcccumulation
6,9642618
answered Jan 22 at 22:29
AcccumulationAcccumulation
6,9642618
answered Jan 22 at 22:29
AcccumulationAcccumulation
6,9642618
answered Jan 22 at 22:29
AcccumulationAcccumulation
6,9642618
6,9642618
add a comment |
add a comment |
$begingroup$
Yes, this is correct. It is because the logarithm function is defined in that manner. A logarithm is essentially asking the question To what power $x$ do I raise some number $a$ to get $b$?
In general, the statements $a^x=b$ and $log_ab=x$ are equivalent.
In this case, since the base of the logarithm is $e$, you can ask yourself the same question: To what power do I raise $e$ to get $e^x$? Clearly the answer is $x$.
I recommend you read up on logarithms if you are having difficulty in this basic concept: https://en.m.wikipedia.org/wiki/Logarithm
answered Jan 22 at 12:54
Naman KumarNaman Kumar
19613
$endgroup$
add a comment |
$begingroup$
Yes, this is correct. It is because the logarithm function is defined in that manner. A logarithm is essentially asking the question To what power $x$ do I raise some number $a$ to get $b$?
In general, the statements $a^x=b$ and $log_ab=x$ are equivalent.
In this case, since the base of the logarithm is $e$, you can ask yourself the same question: To what power do I raise $e$ to get $e^x$? Clearly the answer is $x$.
I recommend you read up on logarithms if you are having difficulty in this basic concept: https://en.m.wikipedia.org/wiki/Logarithm
answered Jan 22 at 12:54
Naman KumarNaman Kumar
19613
$endgroup$
add a comment |
$begingroup$
Yes, this is correct. It is because the logarithm function is defined in that manner. A logarithm is essentially asking the question To what power $x$ do I raise some number $a$ to get $b$?
In general, the statements $a^x=b$ and $log_ab=x$ are equivalent.
In this case, since the base of the logarithm is $e$, you can ask yourself the same question: To what power do I raise $e$ to get $e^x$? Clearly the answer is $x$.
I recommend you read up on logarithms if you are having difficulty in this basic concept: https://en.m.wikipedia.org/wiki/Logarithm
answered Jan 22 at 12:54
Naman KumarNaman Kumar
19613
$endgroup$
Yes, this is correct. It is because the logarithm function is defined in that manner. A logarithm is essentially asking the question To what power $x$ do I raise some number $a$ to get $b$?
In general, the statements $a^x=b$ and $log_ab=x$ are equivalent.
In this case, since the base of the logarithm is $e$, you can ask yourself the same question: To what power do I raise $e$ to get $e^x$? Clearly the answer is $x$.
I recommend you read up on logarithms if you are having difficulty in this basic concept: https://en.m.wikipedia.org/wiki/Logarithm
answered Jan 22 at 12:54
Naman KumarNaman Kumar
19613
answered Jan 22 at 12:54
Naman KumarNaman Kumar
19613
answered Jan 22 at 12:54
Naman KumarNaman Kumar
19613
answered Jan 22 at 12:54
Naman KumarNaman Kumar
19613
19613
add a comment |
add a comment |
$begingroup$
With $x=ln(y)$, $$e^{ln(y)}=yimplies ln(e^{ln(y)})=ln(y)implies ln(e^x)=x.$$
There is no need to refer to inversion nor any other property than the given.
answered Jan 22 at 12:58
Yves DaoustYves Daoust
127k673226
$endgroup$
1
$begingroup$
Haven't you used the following property without stating it? That is: for all x there exists y such x = ln(y); i.e. the logarithm function is onto the set in question (we can probably assume that set is the reals, although it hasn't been stated).
$endgroup$
– Don Hatch
Jan 22 at 22:25
$begingroup$
@DonHatch indeed, but do recall a function is invertible iff it is bijective. In order to talk of the logarithm as an inverse we should probably have valided the exponential is bijective on the reals. Over the complex numbers we lose injectivity, but preserve surjectivity, so we can use the concept of covering spaces or the like to define the "inverse" of a non-injective function.
$endgroup$
– Brevan Ellefsen
Jan 23 at 2:49
$begingroup$
@DonHatch: no, I cheated by saying "take $x=ln y$", so the proof is valid for all such pairs (if any), but only those. Anyway, if we add that the range of the logarithm is $mathbb R$, the proof holds for all real $x$. The range condition is sufficient, invertibility is not required.
$endgroup$
– Yves Daoust
Jan 23 at 9:13
add a comment |
$begingroup$
With $x=ln(y)$, $$e^{ln(y)}=yimplies ln(e^{ln(y)})=ln(y)implies ln(e^x)=x.$$
There is no need to refer to inversion nor any other property than the given.
answered Jan 22 at 12:58
Yves DaoustYves Daoust
127k673226
$endgroup$
1
$begingroup$
Haven't you used the following property without stating it? That is: for all x there exists y such x = ln(y); i.e. the logarithm function is onto the set in question (we can probably assume that set is the reals, although it hasn't been stated).
$endgroup$
– Don Hatch
Jan 22 at 22:25
$begingroup$
@DonHatch indeed, but do recall a function is invertible iff it is bijective. In order to talk of the logarithm as an inverse we should probably have valided the exponential is bijective on the reals. Over the complex numbers we lose injectivity, but preserve surjectivity, so we can use the concept of covering spaces or the like to define the "inverse" of a non-injective function.
$endgroup$
– Brevan Ellefsen
Jan 23 at 2:49
$begingroup$
@DonHatch: no, I cheated by saying "take $x=ln y$", so the proof is valid for all such pairs (if any), but only those. Anyway, if we add that the range of the logarithm is $mathbb R$, the proof holds for all real $x$. The range condition is sufficient, invertibility is not required.
$endgroup$
– Yves Daoust
Jan 23 at 9:13
add a comment |
$begingroup$
With $x=ln(y)$, $$e^{ln(y)}=yimplies ln(e^{ln(y)})=ln(y)implies ln(e^x)=x.$$
There is no need to refer to inversion nor any other property than the given.
answered Jan 22 at 12:58
Yves DaoustYves Daoust
127k673226
$endgroup$
With $x=ln(y)$, $$e^{ln(y)}=yimplies ln(e^{ln(y)})=ln(y)implies ln(e^x)=x.$$
There is no need to refer to inversion nor any other property than the given.
answered Jan 22 at 12:58
Yves DaoustYves Daoust
127k673226
edited Jan 22 at 13:04
answered Jan 22 at 12:58
Yves DaoustYves Daoust
127k673226
answered Jan 22 at 12:58
Yves DaoustYves Daoust
127k673226
answered Jan 22 at 12:58
Yves DaoustYves Daoust
127k673226
127k673226
1
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Haven't you used the following property without stating it? That is: for all x there exists y such x = ln(y); i.e. the logarithm function is onto the set in question (we can probably assume that set is the reals, although it hasn't been stated).
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– Don Hatch
Jan 22 at 22:25
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@DonHatch indeed, but do recall a function is invertible iff it is bijective. In order to talk of the logarithm as an inverse we should probably have valided the exponential is bijective on the reals. Over the complex numbers we lose injectivity, but preserve surjectivity, so we can use the concept of covering spaces or the like to define the "inverse" of a non-injective function.
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– Brevan Ellefsen
Jan 23 at 2:49
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@DonHatch: no, I cheated by saying "take $x=ln y$", so the proof is valid for all such pairs (if any), but only those. Anyway, if we add that the range of the logarithm is $mathbb R$, the proof holds for all real $x$. The range condition is sufficient, invertibility is not required.
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– Yves Daoust
Jan 23 at 9:13
add a comment |
1
$begingroup$
Haven't you used the following property without stating it? That is: for all x there exists y such x = ln(y); i.e. the logarithm function is onto the set in question (we can probably assume that set is the reals, although it hasn't been stated).
$endgroup$
– Don Hatch
Jan 22 at 22:25
$begingroup$
@DonHatch indeed, but do recall a function is invertible iff it is bijective. In order to talk of the logarithm as an inverse we should probably have valided the exponential is bijective on the reals. Over the complex numbers we lose injectivity, but preserve surjectivity, so we can use the concept of covering spaces or the like to define the "inverse" of a non-injective function.
$endgroup$
– Brevan Ellefsen
Jan 23 at 2:49
$begingroup$
@DonHatch: no, I cheated by saying "take $x=ln y$", so the proof is valid for all such pairs (if any), but only those. Anyway, if we add that the range of the logarithm is $mathbb R$, the proof holds for all real $x$. The range condition is sufficient, invertibility is not required.
$endgroup$
– Yves Daoust
Jan 23 at 9:13
1
1
$begingroup$
Haven't you used the following property without stating it? That is: for all x there exists y such x = ln(y); i.e. the logarithm function is onto the set in question (we can probably assume that set is the reals, although it hasn't been stated).
$endgroup$
– Don Hatch
Jan 22 at 22:25
$begingroup$
Haven't you used the following property without stating it? That is: for all x there exists y such x = ln(y); i.e. the logarithm function is onto the set in question (we can probably assume that set is the reals, although it hasn't been stated).
$endgroup$
– Don Hatch
Jan 22 at 22:25
$begingroup$
@DonHatch indeed, but do recall a function is invertible iff it is bijective. In order to talk of the logarithm as an inverse we should probably have valided the exponential is bijective on the reals. Over the complex numbers we lose injectivity, but preserve surjectivity, so we can use the concept of covering spaces or the like to define the "inverse" of a non-injective function.
$endgroup$
– Brevan Ellefsen
Jan 23 at 2:49
$begingroup$
@DonHatch indeed, but do recall a function is invertible iff it is bijective. In order to talk of the logarithm as an inverse we should probably have valided the exponential is bijective on the reals. Over the complex numbers we lose injectivity, but preserve surjectivity, so we can use the concept of covering spaces or the like to define the "inverse" of a non-injective function.
$endgroup$
– Brevan Ellefsen
Jan 23 at 2:49
$begingroup$
@DonHatch: no, I cheated by saying "take $x=ln y$", so the proof is valid for all such pairs (if any), but only those. Anyway, if we add that the range of the logarithm is $mathbb R$, the proof holds for all real $x$. The range condition is sufficient, invertibility is not required.
$endgroup$
– Yves Daoust
Jan 23 at 9:13
$begingroup$
@DonHatch: no, I cheated by saying "take $x=ln y$", so the proof is valid for all such pairs (if any), but only those. Anyway, if we add that the range of the logarithm is $mathbb R$, the proof holds for all real $x$. The range condition is sufficient, invertibility is not required.
$endgroup$
– Yves Daoust
Jan 23 at 9:13
add a comment |
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6
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It's true, and the simple explanation is that ln and $e^x$ are each other's inverse functions. And there's no range problem because $e^x > 0$. Only in calculating the latter ($e^{ln{x}}$) you have to take $x>0$.
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– Matti P.
Jan 22 at 12:49
3
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Let's say $ln(e^x)=y$. Then $e^{ln(e^x)}=e^x=e^y$. Here I used your second line. Since the exponential function is injective it follows that $x=y$.
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– maxmilgram
Jan 22 at 12:55
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The history is worth considering: Natural log predated Euler's $e^x$
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– TurlocTheRed
Jan 22 at 22:24
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Are you assuming that $x$ is a real number?
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– Dan
Jan 22 at 23:43
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It is true iff $Im xin(-pi,pi]$. Assuming the branch of $ln$ is principal.
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– Kemono Chen
Jan 23 at 2:13