Don't understand how log rules are applied in formula
0
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On page 148 of Computational Complexity by Papadimitriou, it states:
$nc_1^{f(n)} = c_1^{log(n) + f(n)}$
I understand that I can expand the right part of the formula to $c_1^{log(n) + f(n)} = c_1^{log(n)}c_1^{f(n)}$, but then why and how does $c_1^{log(n)} = n$?
logarithms computational-complexity
edited Jan 14 at 15:15
user376343
3,4383827
asked Jan 14 at 15:14
NateNate
1275
$endgroup$
add a comment |
0
$begingroup$
On page 148 of Computational Complexity by Papadimitriou, it states:
$nc_1^{f(n)} = c_1^{log(n) + f(n)}$
I understand that I can expand the right part of the formula to $c_1^{log(n) + f(n)} = c_1^{log(n)}c_1^{f(n)}$, but then why and how does $c_1^{log(n)} = n$?
logarithms computational-complexity
edited Jan 14 at 15:15
user376343
3,4383827
asked Jan 14 at 15:14
NateNate
1275
$endgroup$
$begingroup$
Just divide both sides by $c_1^{f(n)}$.
$endgroup$
– KM101
Jan 14 at 15:15
2
$begingroup$
Iff the $log(n)$ is taken to base $c_1$, i.e. $log_{c_1}(n)$,then the equality holds.
$endgroup$
– Andreas
Jan 14 at 15:17
$begingroup$
I assumed the book was using $log(n)$ to mean $log_2(n)$, but it's possible the author wasn't.
$endgroup$
– Nate
Jan 14 at 15:57
$begingroup$
Literally, $c_1^{log(n)} = n^{log(c_1)}.$ But $n = c_1^{log(n)/log(c_1)}.$ Maybe something written before or after this makes it work out or gives a hint toward the missing piece. For example, $log(n)/log(c_1) + f(n)$ is $O(log(n) + f(n)).$
$endgroup$
– David K
Jan 14 at 16:17
add a comment |
0
0
0
$begingroup$
On page 148 of Computational Complexity by Papadimitriou, it states:
$nc_1^{f(n)} = c_1^{log(n) + f(n)}$
I understand that I can expand the right part of the formula to $c_1^{log(n) + f(n)} = c_1^{log(n)}c_1^{f(n)}$, but then why and how does $c_1^{log(n)} = n$?
logarithms computational-complexity
edited Jan 14 at 15:15
user376343
3,4383827
asked Jan 14 at 15:14
NateNate
1275
$endgroup$
On page 148 of Computational Complexity by Papadimitriou, it states:
$nc_1^{f(n)} = c_1^{log(n) + f(n)}$
I understand that I can expand the right part of the formula to $c_1^{log(n) + f(n)} = c_1^{log(n)}c_1^{f(n)}$, but then why and how does $c_1^{log(n)} = n$?
logarithms computational-complexity
logarithms computational-complexity
edited Jan 14 at 15:15
user376343
3,4383827
asked Jan 14 at 15:14
NateNate
1275
edited Jan 14 at 15:15
user376343
3,4383827
asked Jan 14 at 15:14
NateNate
1275
edited Jan 14 at 15:15
user376343
3,4383827
edited Jan 14 at 15:15
user376343
3,4383827
edited Jan 14 at 15:15
user376343
3,4383827
3,4383827
asked Jan 14 at 15:14
NateNate
1275
asked Jan 14 at 15:14
NateNate
1275
asked Jan 14 at 15:14
NateNate
1275
1275
$begingroup$
Just divide both sides by $c_1^{f(n)}$.
$endgroup$
– KM101
Jan 14 at 15:15
2
$begingroup$
Iff the $log(n)$ is taken to base $c_1$, i.e. $log_{c_1}(n)$,then the equality holds.
$endgroup$
– Andreas
Jan 14 at 15:17
$begingroup$
I assumed the book was using $log(n)$ to mean $log_2(n)$, but it's possible the author wasn't.
$endgroup$
– Nate
Jan 14 at 15:57
$begingroup$
Literally, $c_1^{log(n)} = n^{log(c_1)}.$ But $n = c_1^{log(n)/log(c_1)}.$ Maybe something written before or after this makes it work out or gives a hint toward the missing piece. For example, $log(n)/log(c_1) + f(n)$ is $O(log(n) + f(n)).$
$endgroup$
– David K
Jan 14 at 16:17
add a comment |
$begingroup$
Just divide both sides by $c_1^{f(n)}$.
$endgroup$
– KM101
Jan 14 at 15:15
2
$begingroup$
Iff the $log(n)$ is taken to base $c_1$, i.e. $log_{c_1}(n)$,then the equality holds.
$endgroup$
– Andreas
Jan 14 at 15:17
$begingroup$
I assumed the book was using $log(n)$ to mean $log_2(n)$, but it's possible the author wasn't.
$endgroup$
– Nate
Jan 14 at 15:57
$begingroup$
Literally, $c_1^{log(n)} = n^{log(c_1)}.$ But $n = c_1^{log(n)/log(c_1)}.$ Maybe something written before or after this makes it work out or gives a hint toward the missing piece. For example, $log(n)/log(c_1) + f(n)$ is $O(log(n) + f(n)).$
$endgroup$
– David K
Jan 14 at 16:17
$begingroup$
Just divide both sides by $c_1^{f(n)}$.
$endgroup$
– KM101
Jan 14 at 15:15
$begingroup$
Just divide both sides by $c_1^{f(n)}$.
$endgroup$
– KM101
Jan 14 at 15:15
2
2
$begingroup$
Iff the $log(n)$ is taken to base $c_1$, i.e. $log_{c_1}(n)$,then the equality holds.
$endgroup$
– Andreas
Jan 14 at 15:17
$begingroup$
Iff the $log(n)$ is taken to base $c_1$, i.e. $log_{c_1}(n)$,then the equality holds.
$endgroup$
– Andreas
Jan 14 at 15:17
$begingroup$
I assumed the book was using $log(n)$ to mean $log_2(n)$, but it's possible the author wasn't.
$endgroup$
– Nate
Jan 14 at 15:57
$begingroup$
I assumed the book was using $log(n)$ to mean $log_2(n)$, but it's possible the author wasn't.
$endgroup$
– Nate
Jan 14 at 15:57
$begingroup$
Literally, $c_1^{log(n)} = n^{log(c_1)}.$ But $n = c_1^{log(n)/log(c_1)}.$ Maybe something written before or after this makes it work out or gives a hint toward the missing piece. For example, $log(n)/log(c_1) + f(n)$ is $O(log(n) + f(n)).$
$endgroup$
– David K
Jan 14 at 16:17
$begingroup$
Literally, $c_1^{log(n)} = n^{log(c_1)}.$ But $n = c_1^{log(n)/log(c_1)}.$ Maybe something written before or after this makes it work out or gives a hint toward the missing piece. For example, $log(n)/log(c_1) + f(n)$ is $O(log(n) + f(n)).$
$endgroup$
– David K
Jan 14 at 16:17
add a comment |
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$begingroup$
Just divide both sides by $c_1^{f(n)}$.
$endgroup$
– KM101
Jan 14 at 15:15
2
$begingroup$
Iff the $log(n)$ is taken to base $c_1$, i.e. $log_{c_1}(n)$,then the equality holds.
$endgroup$
– Andreas
Jan 14 at 15:17
$begingroup$
I assumed the book was using $log(n)$ to mean $log_2(n)$, but it's possible the author wasn't.
$endgroup$
– Nate
Jan 14 at 15:57
$begingroup$
Literally, $c_1^{log(n)} = n^{log(c_1)}.$ But $n = c_1^{log(n)/log(c_1)}.$ Maybe something written before or after this makes it work out or gives a hint toward the missing piece. For example, $log(n)/log(c_1) + f(n)$ is $O(log(n) + f(n)).$
$endgroup$
– David K
Jan 14 at 16:17