Logarithms in Summations : Confusion!
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I see this simplification and I am confused! I thought there is no explicit way to simplify the logarithm of a summation.
Can someone explain how the the second term( involving the summation), gets converted to a log of a summation of exponential log sums ?
The equation is in the image below! Thanks! :)
enter image description here
logarithms
asked Nov 26 '18 at 13:33
Daniel DsouzaDaniel Dsouza
1
$endgroup$
add a comment |
$begingroup$
I see this simplification and I am confused! I thought there is no explicit way to simplify the logarithm of a summation.
Can someone explain how the the second term( involving the summation), gets converted to a log of a summation of exponential log sums ?
The equation is in the image below! Thanks! :)
enter image description here
logarithms
asked Nov 26 '18 at 13:33
Daniel DsouzaDaniel Dsouza
1
$endgroup$
$begingroup$
$e^{(log{a}+log{b})}=e^{log{a}}e^{log{b}}=ab$
$endgroup$
– saulspatz
Nov 26 '18 at 13:36
add a comment |
$begingroup$
I see this simplification and I am confused! I thought there is no explicit way to simplify the logarithm of a summation.
Can someone explain how the the second term( involving the summation), gets converted to a log of a summation of exponential log sums ?
The equation is in the image below! Thanks! :)
enter image description here
logarithms
asked Nov 26 '18 at 13:33
Daniel DsouzaDaniel Dsouza
1
$endgroup$
I see this simplification and I am confused! I thought there is no explicit way to simplify the logarithm of a summation.
Can someone explain how the the second term( involving the summation), gets converted to a log of a summation of exponential log sums ?
The equation is in the image below! Thanks! :)
enter image description here
logarithms
logarithms
asked Nov 26 '18 at 13:33
Daniel DsouzaDaniel Dsouza
1
asked Nov 26 '18 at 13:33
Daniel DsouzaDaniel Dsouza
1
asked Nov 26 '18 at 13:33
Daniel DsouzaDaniel Dsouza
1
asked Nov 26 '18 at 13:33
Daniel DsouzaDaniel Dsouza
1
asked Nov 26 '18 at 13:33
Daniel DsouzaDaniel Dsouza
1
1
$begingroup$
$e^{(log{a}+log{b})}=e^{log{a}}e^{log{b}}=ab$
$endgroup$
– saulspatz
Nov 26 '18 at 13:36
add a comment |
$begingroup$
$e^{(log{a}+log{b})}=e^{log{a}}e^{log{b}}=ab$
$endgroup$
– saulspatz
Nov 26 '18 at 13:36
$begingroup$
$e^{(log{a}+log{b})}=e^{log{a}}e^{log{b}}=ab$
$endgroup$
– saulspatz
Nov 26 '18 at 13:36
$begingroup$
$e^{(log{a}+log{b})}=e^{log{a}}e^{log{b}}=ab$
$endgroup$
– saulspatz
Nov 26 '18 at 13:36
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The main trick is as follows:
begin{align}
a_ib_i &= expleft(log(a_ib_i)right) \
&= exp left( log(a_i) + log(b_i) right)
end{align}
if $a_i, b_i >0$.
answered Nov 26 '18 at 13:37
Siong Thye GohSiong Thye Goh
100k1466117
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add a comment |
$begingroup$
This is just an application of the fact that $log(a)+log(b) = log(ab)$.
answered Nov 26 '18 at 13:35
user3482749user3482749
4,047818
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add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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votes
$begingroup$
The main trick is as follows:
begin{align}
a_ib_i &= expleft(log(a_ib_i)right) \
&= exp left( log(a_i) + log(b_i) right)
end{align}
if $a_i, b_i >0$.
answered Nov 26 '18 at 13:37
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
add a comment |
$begingroup$
The main trick is as follows:
begin{align}
a_ib_i &= expleft(log(a_ib_i)right) \
&= exp left( log(a_i) + log(b_i) right)
end{align}
if $a_i, b_i >0$.
answered Nov 26 '18 at 13:37
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
add a comment |
$begingroup$
The main trick is as follows:
begin{align}
a_ib_i &= expleft(log(a_ib_i)right) \
&= exp left( log(a_i) + log(b_i) right)
end{align}
if $a_i, b_i >0$.
answered Nov 26 '18 at 13:37
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
The main trick is as follows:
begin{align}
a_ib_i &= expleft(log(a_ib_i)right) \
&= exp left( log(a_i) + log(b_i) right)
end{align}
if $a_i, b_i >0$.
answered Nov 26 '18 at 13:37
Siong Thye GohSiong Thye Goh
100k1466117
answered Nov 26 '18 at 13:37
Siong Thye GohSiong Thye Goh
100k1466117
answered Nov 26 '18 at 13:37
Siong Thye GohSiong Thye Goh
100k1466117
answered Nov 26 '18 at 13:37
Siong Thye GohSiong Thye Goh
100k1466117
100k1466117
add a comment |
add a comment |
$begingroup$
This is just an application of the fact that $log(a)+log(b) = log(ab)$.
answered Nov 26 '18 at 13:35
user3482749user3482749
4,047818
$endgroup$
add a comment |
$begingroup$
This is just an application of the fact that $log(a)+log(b) = log(ab)$.
answered Nov 26 '18 at 13:35
user3482749user3482749
4,047818
$endgroup$
add a comment |
$begingroup$
This is just an application of the fact that $log(a)+log(b) = log(ab)$.
answered Nov 26 '18 at 13:35
user3482749user3482749
4,047818
$endgroup$
This is just an application of the fact that $log(a)+log(b) = log(ab)$.
answered Nov 26 '18 at 13:35
user3482749user3482749
4,047818
answered Nov 26 '18 at 13:35
user3482749user3482749
4,047818
answered Nov 26 '18 at 13:35
user3482749user3482749
4,047818
answered Nov 26 '18 at 13:35
user3482749user3482749
4,047818
4,047818
add a comment |
add a comment |
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$begingroup$
$e^{(log{a}+log{b})}=e^{log{a}}e^{log{b}}=ab$
$endgroup$
– saulspatz
Nov 26 '18 at 13:36