Why is $ a^x = e^{x log a} $?
$begingroup$
Why is $ a^x = e^{x log a}$, where $ a $ is a constant?
From my research, I understand that the the natural log of a number is the constant you get when you differentiate the function of that number raised to the power $ x$. For example, if we differentiate the function $ 2^x $, we get $ 2 log 2$. I also understand that the natural log of e is just 1. But I cannot connect the dots here.
I would really appreciate an intuitive explanation of why we can write a number raised to a power as e raised to (the power x the natural logarithm of the number)?
logarithms exponentiation
edited Dec 24 '18 at 14:53
Martín Vacas Vignolo
3,796623
asked Dec 24 '18 at 14:48
WorldGovWorldGov
345211
$endgroup$
add a comment |
$begingroup$
Why is $ a^x = e^{x log a}$, where $ a $ is a constant?
From my research, I understand that the the natural log of a number is the constant you get when you differentiate the function of that number raised to the power $ x$. For example, if we differentiate the function $ 2^x $, we get $ 2 log 2$. I also understand that the natural log of e is just 1. But I cannot connect the dots here.
I would really appreciate an intuitive explanation of why we can write a number raised to a power as e raised to (the power x the natural logarithm of the number)?
logarithms exponentiation
edited Dec 24 '18 at 14:53
Martín Vacas Vignolo
3,796623
asked Dec 24 '18 at 14:48
WorldGovWorldGov
345211
$endgroup$
$begingroup$
In real analysis, $a>0$.
$endgroup$
– Wuestenfux
Dec 24 '18 at 14:49
$begingroup$
That is a definition of $a^x$.
$endgroup$
– Bernard
Dec 24 '18 at 15:36
add a comment |
$begingroup$
Why is $ a^x = e^{x log a}$, where $ a $ is a constant?
From my research, I understand that the the natural log of a number is the constant you get when you differentiate the function of that number raised to the power $ x$. For example, if we differentiate the function $ 2^x $, we get $ 2 log 2$. I also understand that the natural log of e is just 1. But I cannot connect the dots here.
I would really appreciate an intuitive explanation of why we can write a number raised to a power as e raised to (the power x the natural logarithm of the number)?
logarithms exponentiation
edited Dec 24 '18 at 14:53
Martín Vacas Vignolo
3,796623
asked Dec 24 '18 at 14:48
WorldGovWorldGov
345211
$endgroup$
Why is $ a^x = e^{x log a}$, where $ a $ is a constant?
From my research, I understand that the the natural log of a number is the constant you get when you differentiate the function of that number raised to the power $ x$. For example, if we differentiate the function $ 2^x $, we get $ 2 log 2$. I also understand that the natural log of e is just 1. But I cannot connect the dots here.
I would really appreciate an intuitive explanation of why we can write a number raised to a power as e raised to (the power x the natural logarithm of the number)?
logarithms exponentiation
logarithms exponentiation
edited Dec 24 '18 at 14:53
Martín Vacas Vignolo
3,796623
asked Dec 24 '18 at 14:48
WorldGovWorldGov
345211
edited Dec 24 '18 at 14:53
Martín Vacas Vignolo
3,796623
asked Dec 24 '18 at 14:48
WorldGovWorldGov
345211
edited Dec 24 '18 at 14:53
Martín Vacas Vignolo
3,796623
edited Dec 24 '18 at 14:53
Martín Vacas Vignolo
3,796623
edited Dec 24 '18 at 14:53
Martín Vacas Vignolo
3,796623
3,796623
asked Dec 24 '18 at 14:48
WorldGovWorldGov
345211
asked Dec 24 '18 at 14:48
WorldGovWorldGov
345211
asked Dec 24 '18 at 14:48
WorldGovWorldGov
345211
345211
$begingroup$
In real analysis, $a>0$.
$endgroup$
– Wuestenfux
Dec 24 '18 at 14:49
$begingroup$
That is a definition of $a^x$.
$endgroup$
– Bernard
Dec 24 '18 at 15:36
add a comment |
$begingroup$
In real analysis, $a>0$.
$endgroup$
– Wuestenfux
Dec 24 '18 at 14:49
$begingroup$
That is a definition of $a^x$.
$endgroup$
– Bernard
Dec 24 '18 at 15:36
$begingroup$
In real analysis, $a>0$.
$endgroup$
– Wuestenfux
Dec 24 '18 at 14:49
$begingroup$
In real analysis, $a>0$.
$endgroup$
– Wuestenfux
Dec 24 '18 at 14:49
$begingroup$
That is a definition of $a^x$.
$endgroup$
– Bernard
Dec 24 '18 at 15:36
$begingroup$
That is a definition of $a^x$.
$endgroup$
– Bernard
Dec 24 '18 at 15:36
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
I think we can agree that
$$a=e^{log a}$$
which arises from one of the properties of the logarithm. Therefore, it’s sufficient to say that
$$a^x=e^{log a^x}$$
But one of the properties of the logarithm also dictates that
$$log a^x=xlog a$$
Therefore
$$a^x=e^{xlog a}$$
answered Dec 24 '18 at 14:54
Frank W.Frank W.
3,7371321
$endgroup$
add a comment |
$begingroup$
$$e^{xlog a}=e^{log a^x}=a^x$$
answered Dec 24 '18 at 14:54
KKZiomekKKZiomek
2,2441641
$endgroup$
add a comment |
$begingroup$
As noted by the other answers, this is due to $e^x$ and $log x$ being inverses.
However, you can also note that this is a combination of two important properties of logarithms of all bases:
$$log_a b^c = clog_b$$
$$a^{log_a b} = b$$
The second property is easy to understand. $log_a b$ means the number $a$ needs to be raised to to get $b$. So raising $a$ to “the number $a$ needs to be raised to give $b$” obviously gives $b$. Mathematically, you could say
$$log_a b = c iff a^c = b$$
$$a^{log_a b} = a^c = b$$
The first property is can be thought of as a repeated addition property:
$$log_a bc = x iff a^x = bc$$
$x = log_a b+log_a c$ gives
$$underbrace{a^{log_a b+log_a c}}_{a^{log_ ab+log_a c} = a^{log_a b}cdot a^{log_a c} = bc} = bc iff color{blue}{log_a(bc) = log_a b+log_a c}$$
from which you reach the first property.
$$log_a (bc) = log b+log c implies log_a b^c = underbrace{log_a b+log_ab+…+log_a b}_{c text{ times}} = clog_a b$$
Combining the two properties, you get
$$e^{alog x} = e^{log a^x} = a^x$$
answered Dec 24 '18 at 15:23
KM101KM101
6,0901525
$endgroup$
add a comment |
$begingroup$
$ln(x)$ is the inverse function of $e^x$ and so we have $e^{ln(x)}=x$ moreover, from the properties of $ln(x)$ is that $ln(a^b)=bln(a)$ so $a^x=e^{ln(a^x)}=e^{xln(a)}$
$endgroup$
1
$begingroup$
Use a backslash in front of $ln$ to render it properly.
$endgroup$
– Don Thousand
Dec 24 '18 at 14:56
$begingroup$
thank you, didn't know about it
$endgroup$
– user531476
Dec 24 '18 at 14:58
add a comment |
$begingroup$
The decisive point is that the functions $exp:{Bbb R}rightarrow {Bbb R}_{>0}:xmapsto e^x$ and $log:{Bbb R}_{>0}rightarrow {Bbb R}:xmapsto log_ex = ln x$ are inverse to each other. Then we have $a^x = e^{ln a^x}$. By the property of the logarithm, $e^{ln a^x} = e^{xln a}$.
answered Dec 24 '18 at 15:05
WuestenfuxWuestenfux
5,3631513
$endgroup$
add a comment |
$begingroup$
Use that $$ln(e^p)=pimpliesln(e^{ln q})=ln qimplies e^{ln q}=q$$
The first result will take you the rest of the way.
answered Dec 24 '18 at 15:16
Rhys HughesRhys Hughes
7,0501630
$endgroup$
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think we can agree that
$$a=e^{log a}$$
which arises from one of the properties of the logarithm. Therefore, it’s sufficient to say that
$$a^x=e^{log a^x}$$
But one of the properties of the logarithm also dictates that
$$log a^x=xlog a$$
Therefore
$$a^x=e^{xlog a}$$
answered Dec 24 '18 at 14:54
Frank W.Frank W.
3,7371321
$endgroup$
add a comment |
$begingroup$
I think we can agree that
$$a=e^{log a}$$
which arises from one of the properties of the logarithm. Therefore, it’s sufficient to say that
$$a^x=e^{log a^x}$$
But one of the properties of the logarithm also dictates that
$$log a^x=xlog a$$
Therefore
$$a^x=e^{xlog a}$$
answered Dec 24 '18 at 14:54
Frank W.Frank W.
3,7371321
$endgroup$
add a comment |
$begingroup$
I think we can agree that
$$a=e^{log a}$$
which arises from one of the properties of the logarithm. Therefore, it’s sufficient to say that
$$a^x=e^{log a^x}$$
But one of the properties of the logarithm also dictates that
$$log a^x=xlog a$$
Therefore
$$a^x=e^{xlog a}$$
answered Dec 24 '18 at 14:54
Frank W.Frank W.
3,7371321
$endgroup$
I think we can agree that
$$a=e^{log a}$$
which arises from one of the properties of the logarithm. Therefore, it’s sufficient to say that
$$a^x=e^{log a^x}$$
But one of the properties of the logarithm also dictates that
$$log a^x=xlog a$$
Therefore
$$a^x=e^{xlog a}$$
answered Dec 24 '18 at 14:54
Frank W.Frank W.
3,7371321
answered Dec 24 '18 at 14:54
Frank W.Frank W.
3,7371321
answered Dec 24 '18 at 14:54
Frank W.Frank W.
3,7371321
answered Dec 24 '18 at 14:54
Frank W.Frank W.
3,7371321
3,7371321
add a comment |
add a comment |
$begingroup$
$$e^{xlog a}=e^{log a^x}=a^x$$
answered Dec 24 '18 at 14:54
KKZiomekKKZiomek
2,2441641
$endgroup$
add a comment |
$begingroup$
$$e^{xlog a}=e^{log a^x}=a^x$$
answered Dec 24 '18 at 14:54
KKZiomekKKZiomek
2,2441641
$endgroup$
add a comment |
$begingroup$
$$e^{xlog a}=e^{log a^x}=a^x$$
answered Dec 24 '18 at 14:54
KKZiomekKKZiomek
2,2441641
$endgroup$
$$e^{xlog a}=e^{log a^x}=a^x$$
answered Dec 24 '18 at 14:54
KKZiomekKKZiomek
2,2441641
answered Dec 24 '18 at 14:54
KKZiomekKKZiomek
2,2441641
answered Dec 24 '18 at 14:54
KKZiomekKKZiomek
2,2441641
answered Dec 24 '18 at 14:54
KKZiomekKKZiomek
2,2441641
2,2441641
add a comment |
add a comment |
$begingroup$
As noted by the other answers, this is due to $e^x$ and $log x$ being inverses.
However, you can also note that this is a combination of two important properties of logarithms of all bases:
$$log_a b^c = clog_b$$
$$a^{log_a b} = b$$
The second property is easy to understand. $log_a b$ means the number $a$ needs to be raised to to get $b$. So raising $a$ to “the number $a$ needs to be raised to give $b$” obviously gives $b$. Mathematically, you could say
$$log_a b = c iff a^c = b$$
$$a^{log_a b} = a^c = b$$
The first property is can be thought of as a repeated addition property:
$$log_a bc = x iff a^x = bc$$
$x = log_a b+log_a c$ gives
$$underbrace{a^{log_a b+log_a c}}_{a^{log_ ab+log_a c} = a^{log_a b}cdot a^{log_a c} = bc} = bc iff color{blue}{log_a(bc) = log_a b+log_a c}$$
from which you reach the first property.
$$log_a (bc) = log b+log c implies log_a b^c = underbrace{log_a b+log_ab+…+log_a b}_{c text{ times}} = clog_a b$$
Combining the two properties, you get
$$e^{alog x} = e^{log a^x} = a^x$$
answered Dec 24 '18 at 15:23
KM101KM101
6,0901525
$endgroup$
add a comment |
$begingroup$
As noted by the other answers, this is due to $e^x$ and $log x$ being inverses.
However, you can also note that this is a combination of two important properties of logarithms of all bases:
$$log_a b^c = clog_b$$
$$a^{log_a b} = b$$
The second property is easy to understand. $log_a b$ means the number $a$ needs to be raised to to get $b$. So raising $a$ to “the number $a$ needs to be raised to give $b$” obviously gives $b$. Mathematically, you could say
$$log_a b = c iff a^c = b$$
$$a^{log_a b} = a^c = b$$
The first property is can be thought of as a repeated addition property:
$$log_a bc = x iff a^x = bc$$
$x = log_a b+log_a c$ gives
$$underbrace{a^{log_a b+log_a c}}_{a^{log_ ab+log_a c} = a^{log_a b}cdot a^{log_a c} = bc} = bc iff color{blue}{log_a(bc) = log_a b+log_a c}$$
from which you reach the first property.
$$log_a (bc) = log b+log c implies log_a b^c = underbrace{log_a b+log_ab+…+log_a b}_{c text{ times}} = clog_a b$$
Combining the two properties, you get
$$e^{alog x} = e^{log a^x} = a^x$$
answered Dec 24 '18 at 15:23
KM101KM101
6,0901525
$endgroup$
add a comment |
$begingroup$
As noted by the other answers, this is due to $e^x$ and $log x$ being inverses.
However, you can also note that this is a combination of two important properties of logarithms of all bases:
$$log_a b^c = clog_b$$
$$a^{log_a b} = b$$
The second property is easy to understand. $log_a b$ means the number $a$ needs to be raised to to get $b$. So raising $a$ to “the number $a$ needs to be raised to give $b$” obviously gives $b$. Mathematically, you could say
$$log_a b = c iff a^c = b$$
$$a^{log_a b} = a^c = b$$
The first property is can be thought of as a repeated addition property:
$$log_a bc = x iff a^x = bc$$
$x = log_a b+log_a c$ gives
$$underbrace{a^{log_a b+log_a c}}_{a^{log_ ab+log_a c} = a^{log_a b}cdot a^{log_a c} = bc} = bc iff color{blue}{log_a(bc) = log_a b+log_a c}$$
from which you reach the first property.
$$log_a (bc) = log b+log c implies log_a b^c = underbrace{log_a b+log_ab+…+log_a b}_{c text{ times}} = clog_a b$$
Combining the two properties, you get
$$e^{alog x} = e^{log a^x} = a^x$$
answered Dec 24 '18 at 15:23
KM101KM101
6,0901525
$endgroup$
As noted by the other answers, this is due to $e^x$ and $log x$ being inverses.
However, you can also note that this is a combination of two important properties of logarithms of all bases:
$$log_a b^c = clog_b$$
$$a^{log_a b} = b$$
The second property is easy to understand. $log_a b$ means the number $a$ needs to be raised to to get $b$. So raising $a$ to “the number $a$ needs to be raised to give $b$” obviously gives $b$. Mathematically, you could say
$$log_a b = c iff a^c = b$$
$$a^{log_a b} = a^c = b$$
The first property is can be thought of as a repeated addition property:
$$log_a bc = x iff a^x = bc$$
$x = log_a b+log_a c$ gives
$$underbrace{a^{log_a b+log_a c}}_{a^{log_ ab+log_a c} = a^{log_a b}cdot a^{log_a c} = bc} = bc iff color{blue}{log_a(bc) = log_a b+log_a c}$$
from which you reach the first property.
$$log_a (bc) = log b+log c implies log_a b^c = underbrace{log_a b+log_ab+…+log_a b}_{c text{ times}} = clog_a b$$
Combining the two properties, you get
$$e^{alog x} = e^{log a^x} = a^x$$
answered Dec 24 '18 at 15:23
KM101KM101
6,0901525
edited Dec 24 '18 at 15:42
answered Dec 24 '18 at 15:23
KM101KM101
6,0901525
answered Dec 24 '18 at 15:23
KM101KM101
6,0901525
answered Dec 24 '18 at 15:23
KM101KM101
6,0901525
6,0901525
add a comment |
add a comment |
$begingroup$
$ln(x)$ is the inverse function of $e^x$ and so we have $e^{ln(x)}=x$ moreover, from the properties of $ln(x)$ is that $ln(a^b)=bln(a)$ so $a^x=e^{ln(a^x)}=e^{xln(a)}$
$endgroup$
1
$begingroup$
Use a backslash in front of $ln$ to render it properly.
$endgroup$
– Don Thousand
Dec 24 '18 at 14:56
$begingroup$
thank you, didn't know about it
$endgroup$
– user531476
Dec 24 '18 at 14:58
add a comment |
$begingroup$
$ln(x)$ is the inverse function of $e^x$ and so we have $e^{ln(x)}=x$ moreover, from the properties of $ln(x)$ is that $ln(a^b)=bln(a)$ so $a^x=e^{ln(a^x)}=e^{xln(a)}$
$endgroup$
1
$begingroup$
Use a backslash in front of $ln$ to render it properly.
$endgroup$
– Don Thousand
Dec 24 '18 at 14:56
$begingroup$
thank you, didn't know about it
$endgroup$
– user531476
Dec 24 '18 at 14:58
add a comment |
$begingroup$
$ln(x)$ is the inverse function of $e^x$ and so we have $e^{ln(x)}=x$ moreover, from the properties of $ln(x)$ is that $ln(a^b)=bln(a)$ so $a^x=e^{ln(a^x)}=e^{xln(a)}$
$endgroup$
$ln(x)$ is the inverse function of $e^x$ and so we have $e^{ln(x)}=x$ moreover, from the properties of $ln(x)$ is that $ln(a^b)=bln(a)$ so $a^x=e^{ln(a^x)}=e^{xln(a)}$
edited Dec 24 '18 at 14:58
answered Dec 24 '18 at 14:55
user531476
1
$begingroup$
Use a backslash in front of $ln$ to render it properly.
$endgroup$
– Don Thousand
Dec 24 '18 at 14:56
$begingroup$
thank you, didn't know about it
$endgroup$
– user531476
Dec 24 '18 at 14:58
add a comment |
1
$begingroup$
Use a backslash in front of $ln$ to render it properly.
$endgroup$
– Don Thousand
Dec 24 '18 at 14:56
$begingroup$
thank you, didn't know about it
$endgroup$
– user531476
Dec 24 '18 at 14:58
1
1
$begingroup$
Use a backslash in front of $ln$ to render it properly.
$endgroup$
– Don Thousand
Dec 24 '18 at 14:56
$begingroup$
Use a backslash in front of $ln$ to render it properly.
$endgroup$
– Don Thousand
Dec 24 '18 at 14:56
$begingroup$
thank you, didn't know about it
$endgroup$
– user531476
Dec 24 '18 at 14:58
$begingroup$
thank you, didn't know about it
$endgroup$
– user531476
Dec 24 '18 at 14:58
add a comment |
$begingroup$
The decisive point is that the functions $exp:{Bbb R}rightarrow {Bbb R}_{>0}:xmapsto e^x$ and $log:{Bbb R}_{>0}rightarrow {Bbb R}:xmapsto log_ex = ln x$ are inverse to each other. Then we have $a^x = e^{ln a^x}$. By the property of the logarithm, $e^{ln a^x} = e^{xln a}$.
answered Dec 24 '18 at 15:05
WuestenfuxWuestenfux
5,3631513
$endgroup$
add a comment |
$begingroup$
The decisive point is that the functions $exp:{Bbb R}rightarrow {Bbb R}_{>0}:xmapsto e^x$ and $log:{Bbb R}_{>0}rightarrow {Bbb R}:xmapsto log_ex = ln x$ are inverse to each other. Then we have $a^x = e^{ln a^x}$. By the property of the logarithm, $e^{ln a^x} = e^{xln a}$.
answered Dec 24 '18 at 15:05
WuestenfuxWuestenfux
5,3631513
$endgroup$
add a comment |
$begingroup$
The decisive point is that the functions $exp:{Bbb R}rightarrow {Bbb R}_{>0}:xmapsto e^x$ and $log:{Bbb R}_{>0}rightarrow {Bbb R}:xmapsto log_ex = ln x$ are inverse to each other. Then we have $a^x = e^{ln a^x}$. By the property of the logarithm, $e^{ln a^x} = e^{xln a}$.
answered Dec 24 '18 at 15:05
WuestenfuxWuestenfux
5,3631513
$endgroup$
The decisive point is that the functions $exp:{Bbb R}rightarrow {Bbb R}_{>0}:xmapsto e^x$ and $log:{Bbb R}_{>0}rightarrow {Bbb R}:xmapsto log_ex = ln x$ are inverse to each other. Then we have $a^x = e^{ln a^x}$. By the property of the logarithm, $e^{ln a^x} = e^{xln a}$.
answered Dec 24 '18 at 15:05
WuestenfuxWuestenfux
5,3631513
answered Dec 24 '18 at 15:05
WuestenfuxWuestenfux
5,3631513
answered Dec 24 '18 at 15:05
WuestenfuxWuestenfux
5,3631513
answered Dec 24 '18 at 15:05
WuestenfuxWuestenfux
5,3631513
5,3631513
add a comment |
add a comment |
$begingroup$
Use that $$ln(e^p)=pimpliesln(e^{ln q})=ln qimplies e^{ln q}=q$$
The first result will take you the rest of the way.
answered Dec 24 '18 at 15:16
Rhys HughesRhys Hughes
7,0501630
$endgroup$
add a comment |
$begingroup$
Use that $$ln(e^p)=pimpliesln(e^{ln q})=ln qimplies e^{ln q}=q$$
The first result will take you the rest of the way.
answered Dec 24 '18 at 15:16
Rhys HughesRhys Hughes
7,0501630
$endgroup$
add a comment |
$begingroup$
Use that $$ln(e^p)=pimpliesln(e^{ln q})=ln qimplies e^{ln q}=q$$
The first result will take you the rest of the way.
answered Dec 24 '18 at 15:16
Rhys HughesRhys Hughes
7,0501630
$endgroup$
Use that $$ln(e^p)=pimpliesln(e^{ln q})=ln qimplies e^{ln q}=q$$
The first result will take you the rest of the way.
answered Dec 24 '18 at 15:16
Rhys HughesRhys Hughes
7,0501630
answered Dec 24 '18 at 15:16
Rhys HughesRhys Hughes
7,0501630
answered Dec 24 '18 at 15:16
Rhys HughesRhys Hughes
7,0501630
answered Dec 24 '18 at 15:16
Rhys HughesRhys Hughes
7,0501630
7,0501630
add a comment |
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$begingroup$
In real analysis, $a>0$.
$endgroup$
– Wuestenfux
Dec 24 '18 at 14:49
$begingroup$
That is a definition of $a^x$.
$endgroup$
– Bernard
Dec 24 '18 at 15:36