Is there a law for summing powers of logarithms?

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I have the expression



$$(ln a)^c+(ln b)^c$$



Is there a way to combine these (i.e., remove the $+$ sign) to create a single expression in terms of $a,b,c$?




summation logarithms exponentiation




share|cite|improve this question






















  • Have you tried something?
    – Matti P.
    Aug 24 at 9:05










  • I can't work out what to try. Seems like I could somehow take a $log$ of both expressions and then change the base, but I can't see how.
    – Richard Burke-Ward
    Aug 24 at 9:06











  • If we would have terms like $ ccdot ln(a) $ , we could convert the expression, but this expression cannot be converted.
    – Peter
    Aug 24 at 9:25











  • Logarithms are (or at least want to be) exponents. There is no simple rule to deal with (and transform) something like $e^x^y$ the way you can transform $e^xy$ or $e^x+y$, and thus I wouldn't expect there to be any corresponding logarithm rule (for instance, $e^xy = (e^x)^y$ corresponds to $yln x = ln x^y$).
    – Arthur
    Aug 24 at 9:26















up vote
0
down vote

favorite












I have the expression



$$(ln a)^c+(ln b)^c$$



Is there a way to combine these (i.e., remove the $+$ sign) to create a single expression in terms of $a,b,c$?




summation logarithms exponentiation




share|cite|improve this question






















  • Have you tried something?
    – Matti P.
    Aug 24 at 9:05










  • I can't work out what to try. Seems like I could somehow take a $log$ of both expressions and then change the base, but I can't see how.
    – Richard Burke-Ward
    Aug 24 at 9:06











  • If we would have terms like $ ccdot ln(a) $ , we could convert the expression, but this expression cannot be converted.
    – Peter
    Aug 24 at 9:25











  • Logarithms are (or at least want to be) exponents. There is no simple rule to deal with (and transform) something like $e^x^y$ the way you can transform $e^xy$ or $e^x+y$, and thus I wouldn't expect there to be any corresponding logarithm rule (for instance, $e^xy = (e^x)^y$ corresponds to $yln x = ln x^y$).
    – Arthur
    Aug 24 at 9:26













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have the expression



$$(ln a)^c+(ln b)^c$$



Is there a way to combine these (i.e., remove the $+$ sign) to create a single expression in terms of $a,b,c$?




summation logarithms exponentiation




share|cite|improve this question














I have the expression



$$(ln a)^c+(ln b)^c$$



Is there a way to combine these (i.e., remove the $+$ sign) to create a single expression in terms of $a,b,c$?





summation logarithms exponentiation





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 24 at 9:22

























asked Aug 24 at 9:02









Richard Burke-Ward

1628




1628











  • Have you tried something?
    – Matti P.
    Aug 24 at 9:05










  • I can't work out what to try. Seems like I could somehow take a $log$ of both expressions and then change the base, but I can't see how.
    – Richard Burke-Ward
    Aug 24 at 9:06











  • If we would have terms like $ ccdot ln(a) $ , we could convert the expression, but this expression cannot be converted.
    – Peter
    Aug 24 at 9:25











  • Logarithms are (or at least want to be) exponents. There is no simple rule to deal with (and transform) something like $e^x^y$ the way you can transform $e^xy$ or $e^x+y$, and thus I wouldn't expect there to be any corresponding logarithm rule (for instance, $e^xy = (e^x)^y$ corresponds to $yln x = ln x^y$).
    – Arthur
    Aug 24 at 9:26

















  • Have you tried something?
    – Matti P.
    Aug 24 at 9:05










  • I can't work out what to try. Seems like I could somehow take a $log$ of both expressions and then change the base, but I can't see how.
    – Richard Burke-Ward
    Aug 24 at 9:06











  • If we would have terms like $ ccdot ln(a) $ , we could convert the expression, but this expression cannot be converted.
    – Peter
    Aug 24 at 9:25











  • Logarithms are (or at least want to be) exponents. There is no simple rule to deal with (and transform) something like $e^x^y$ the way you can transform $e^xy$ or $e^x+y$, and thus I wouldn't expect there to be any corresponding logarithm rule (for instance, $e^xy = (e^x)^y$ corresponds to $yln x = ln x^y$).
    – Arthur
    Aug 24 at 9:26
















Have you tried something?
– Matti P.
Aug 24 at 9:05




Have you tried something?
– Matti P.
Aug 24 at 9:05












I can't work out what to try. Seems like I could somehow take a $log$ of both expressions and then change the base, but I can't see how.
– Richard Burke-Ward
Aug 24 at 9:06





I can't work out what to try. Seems like I could somehow take a $log$ of both expressions and then change the base, but I can't see how.
– Richard Burke-Ward
Aug 24 at 9:06













If we would have terms like $ ccdot ln(a) $ , we could convert the expression, but this expression cannot be converted.
– Peter
Aug 24 at 9:25





If we would have terms like $ ccdot ln(a) $ , we could convert the expression, but this expression cannot be converted.
– Peter
Aug 24 at 9:25













Logarithms are (or at least want to be) exponents. There is no simple rule to deal with (and transform) something like $e^x^y$ the way you can transform $e^xy$ or $e^x+y$, and thus I wouldn't expect there to be any corresponding logarithm rule (for instance, $e^xy = (e^x)^y$ corresponds to $yln x = ln x^y$).
– Arthur
Aug 24 at 9:26





Logarithms are (or at least want to be) exponents. There is no simple rule to deal with (and transform) something like $e^x^y$ the way you can transform $e^xy$ or $e^x+y$, and thus I wouldn't expect there to be any corresponding logarithm rule (for instance, $e^xy = (e^x)^y$ corresponds to $yln x = ln x^y$).
– Arthur
Aug 24 at 9:26











1 Answer
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up vote
0
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No :



You could simplify the different $ln (a^c)+ln (b^c)=cln (ab)$ or something similar



but unless you know something special about the relationship between $a, b, c$ then $(ln a)^c+(ln b)^c$ is essentially as difficult to simplify as $2^c+3^c$






share|cite|improve this answer




















  • So, a general rule clearly doesn't exist. But in its absence, the specific function I'm looking at right now is $biggl(ln bigl(cos (ln x)+isin (ln x)bigr)biggr)^c+biggl(ln bigl(cos (ln x)-isin (ln x)bigr)biggr)^c$. Any thoughts?
    – Richard Burke-Ward
    Aug 24 at 12:53











  • @RichardBurke-Ward - That expression introduces the traps of the complex logarithm and powers but there may be possible simplifications using $cos(theta) + i sin(theta) = e^i theta$ so it becomes $left(log_e left(exp (i log_ex)right)right)^c+left(log_eleft(exp(-ilog_ex)right)right)^c$ and that might be something like $left(log_e(x)right)^cleft(i^c+(-i)^cright)$
    – Henry
    Aug 24 at 13:31











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1 Answer
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active

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up vote
0
down vote













No :



You could simplify the different $ln (a^c)+ln (b^c)=cln (ab)$ or something similar



but unless you know something special about the relationship between $a, b, c$ then $(ln a)^c+(ln b)^c$ is essentially as difficult to simplify as $2^c+3^c$






share|cite|improve this answer




















  • So, a general rule clearly doesn't exist. But in its absence, the specific function I'm looking at right now is $biggl(ln bigl(cos (ln x)+isin (ln x)bigr)biggr)^c+biggl(ln bigl(cos (ln x)-isin (ln x)bigr)biggr)^c$. Any thoughts?
    – Richard Burke-Ward
    Aug 24 at 12:53











  • @RichardBurke-Ward - That expression introduces the traps of the complex logarithm and powers but there may be possible simplifications using $cos(theta) + i sin(theta) = e^i theta$ so it becomes $left(log_e left(exp (i log_ex)right)right)^c+left(log_eleft(exp(-ilog_ex)right)right)^c$ and that might be something like $left(log_e(x)right)^cleft(i^c+(-i)^cright)$
    – Henry
    Aug 24 at 13:31















up vote
0
down vote













No :



You could simplify the different $ln (a^c)+ln (b^c)=cln (ab)$ or something similar



but unless you know something special about the relationship between $a, b, c$ then $(ln a)^c+(ln b)^c$ is essentially as difficult to simplify as $2^c+3^c$






share|cite|improve this answer




















  • So, a general rule clearly doesn't exist. But in its absence, the specific function I'm looking at right now is $biggl(ln bigl(cos (ln x)+isin (ln x)bigr)biggr)^c+biggl(ln bigl(cos (ln x)-isin (ln x)bigr)biggr)^c$. Any thoughts?
    – Richard Burke-Ward
    Aug 24 at 12:53











  • @RichardBurke-Ward - That expression introduces the traps of the complex logarithm and powers but there may be possible simplifications using $cos(theta) + i sin(theta) = e^i theta$ so it becomes $left(log_e left(exp (i log_ex)right)right)^c+left(log_eleft(exp(-ilog_ex)right)right)^c$ and that might be something like $left(log_e(x)right)^cleft(i^c+(-i)^cright)$
    – Henry
    Aug 24 at 13:31













up vote
0
down vote










up vote
0
down vote









No :



You could simplify the different $ln (a^c)+ln (b^c)=cln (ab)$ or something similar



but unless you know something special about the relationship between $a, b, c$ then $(ln a)^c+(ln b)^c$ is essentially as difficult to simplify as $2^c+3^c$






share|cite|improve this answer












No :



You could simplify the different $ln (a^c)+ln (b^c)=cln (ab)$ or something similar



but unless you know something special about the relationship between $a, b, c$ then $(ln a)^c+(ln b)^c$ is essentially as difficult to simplify as $2^c+3^c$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 24 at 12:17









Henry

93.6k471149




93.6k471149











  • So, a general rule clearly doesn't exist. But in its absence, the specific function I'm looking at right now is $biggl(ln bigl(cos (ln x)+isin (ln x)bigr)biggr)^c+biggl(ln bigl(cos (ln x)-isin (ln x)bigr)biggr)^c$. Any thoughts?
    – Richard Burke-Ward
    Aug 24 at 12:53











  • @RichardBurke-Ward - That expression introduces the traps of the complex logarithm and powers but there may be possible simplifications using $cos(theta) + i sin(theta) = e^i theta$ so it becomes $left(log_e left(exp (i log_ex)right)right)^c+left(log_eleft(exp(-ilog_ex)right)right)^c$ and that might be something like $left(log_e(x)right)^cleft(i^c+(-i)^cright)$
    – Henry
    Aug 24 at 13:31

















  • So, a general rule clearly doesn't exist. But in its absence, the specific function I'm looking at right now is $biggl(ln bigl(cos (ln x)+isin (ln x)bigr)biggr)^c+biggl(ln bigl(cos (ln x)-isin (ln x)bigr)biggr)^c$. Any thoughts?
    – Richard Burke-Ward
    Aug 24 at 12:53











  • @RichardBurke-Ward - That expression introduces the traps of the complex logarithm and powers but there may be possible simplifications using $cos(theta) + i sin(theta) = e^i theta$ so it becomes $left(log_e left(exp (i log_ex)right)right)^c+left(log_eleft(exp(-ilog_ex)right)right)^c$ and that might be something like $left(log_e(x)right)^cleft(i^c+(-i)^cright)$
    – Henry
    Aug 24 at 13:31
















So, a general rule clearly doesn't exist. But in its absence, the specific function I'm looking at right now is $biggl(ln bigl(cos (ln x)+isin (ln x)bigr)biggr)^c+biggl(ln bigl(cos (ln x)-isin (ln x)bigr)biggr)^c$. Any thoughts?
– Richard Burke-Ward
Aug 24 at 12:53





So, a general rule clearly doesn't exist. But in its absence, the specific function I'm looking at right now is $biggl(ln bigl(cos (ln x)+isin (ln x)bigr)biggr)^c+biggl(ln bigl(cos (ln x)-isin (ln x)bigr)biggr)^c$. Any thoughts?
– Richard Burke-Ward
Aug 24 at 12:53













@RichardBurke-Ward - That expression introduces the traps of the complex logarithm and powers but there may be possible simplifications using $cos(theta) + i sin(theta) = e^i theta$ so it becomes $left(log_e left(exp (i log_ex)right)right)^c+left(log_eleft(exp(-ilog_ex)right)right)^c$ and that might be something like $left(log_e(x)right)^cleft(i^c+(-i)^cright)$
– Henry
Aug 24 at 13:31





@RichardBurke-Ward - That expression introduces the traps of the complex logarithm and powers but there may be possible simplifications using $cos(theta) + i sin(theta) = e^i theta$ so it becomes $left(log_e left(exp (i log_ex)right)right)^c+left(log_eleft(exp(-ilog_ex)right)right)^c$ and that might be something like $left(log_e(x)right)^cleft(i^c+(-i)^cright)$
– Henry
Aug 24 at 13:31


















 

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