Logarithms in Summations : Confusion!
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 Dsouza
1
add a comment |
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 Dsouza
1
$e^{(log{a}+log{b})}=e^{log{a}}e^{log{b}}=ab$
– saulspatz
Nov 26 '18 at 13:36
add a comment |
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 Dsouza
1
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 Dsouza
1
asked Nov 26 '18 at 13:33
Daniel Dsouza
1
asked Nov 26 '18 at 13:33
Daniel Dsouza
1
asked Nov 26 '18 at 13:33
Daniel Dsouza
1
asked Nov 26 '18 at 13:33
Daniel Dsouza
1
1
$e^{(log{a}+log{b})}=e^{log{a}}e^{log{b}}=ab$
– saulspatz
Nov 26 '18 at 13:36
add a comment |
$e^{(log{a}+log{b})}=e^{log{a}}e^{log{b}}=ab$
– saulspatz
Nov 26 '18 at 13:36
$e^{(log{a}+log{b})}=e^{log{a}}e^{log{b}}=ab$
– saulspatz
Nov 26 '18 at 13:36
$e^{(log{a}+log{b})}=e^{log{a}}e^{log{b}}=ab$
– saulspatz
Nov 26 '18 at 13:36
add a comment |
2 Answers
2
active
oldest
votes
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 Goh
99.5k1464117
add a comment |
This is just an application of the fact that $log(a)+log(b) = log(ab)$.
answered Nov 26 '18 at 13:35
user3482749
2,708414
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 Goh
99.5k1464117
add a comment |
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 Goh
99.5k1464117
add a comment |
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 Goh
99.5k1464117
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 Goh
99.5k1464117
answered Nov 26 '18 at 13:37
Siong Thye Goh
99.5k1464117
answered Nov 26 '18 at 13:37
Siong Thye Goh
99.5k1464117
answered Nov 26 '18 at 13:37
Siong Thye Goh
99.5k1464117
99.5k1464117
add a comment |
add a comment |
This is just an application of the fact that $log(a)+log(b) = log(ab)$.
answered Nov 26 '18 at 13:35
user3482749
2,708414
add a comment |
This is just an application of the fact that $log(a)+log(b) = log(ab)$.
answered Nov 26 '18 at 13:35
user3482749
2,708414
add a comment |
This is just an application of the fact that $log(a)+log(b) = log(ab)$.
answered Nov 26 '18 at 13:35
user3482749
2,708414
This is just an application of the fact that $log(a)+log(b) = log(ab)$.
answered Nov 26 '18 at 13:35
user3482749
2,708414
answered Nov 26 '18 at 13:35
user3482749
2,708414
answered Nov 26 '18 at 13:35
user3482749
2,708414
answered Nov 26 '18 at 13:35
user3482749
2,708414
2,708414
add a comment |
add a comment |
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$e^{(log{a}+log{b})}=e^{log{a}}e^{log{b}}=ab$
– saulspatz
Nov 26 '18 at 13:36