mersenneforum.org An algebraic quandry
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 2010-07-08, 15:54 #1 Unregistered   5·7·37 Posts An algebraic quandry Hello all, I am a tutor with a student who was a professor with a habit of assigning very challenging derivative problems for extra credit. The last two I have helped this student with are well over 40-50 steps to actually solve. On the current problem, which I will not type out because it would take forever to get the notation and parentheses and exponents correct, I have hit a bit of a wall on simplifying. Note: I can still find the solution but it would be easier if I can implement a different strategy. My question is this: I know logarithms of like bases can be combined via VERY basic log rules -aka --- log(x) + log(x) = log(x^2) = 2 log(x) My student knows this as well. However, my question is this: Is there any relationship between more complicated expressions such as: log(x)*log(x) + log(x) which does NOT equal 2*(log(x))^2 If so, can this be extended to even more complicated expressions such as: log(A)*log(B) + log(C) where A, B, and C are all linear or exponential expressions involving the independent variable x. Note- I do not care how ugly the expression gets- I rather like ugly expressions. However, my student's knowledge is limited to Calc 2, mine is limited to ODE Thanks!
2010-07-08, 16:53   #2
Orgasmic Troll
Cranksta Rap Ayatollah

Jul 2003

641 Posts

Quote:
 Originally Posted by Unregistered Hello all, I am a tutor with a student who was a professor with a habit of assigning very challenging derivative problems for extra credit. The last two I have helped this student with are well over 40-50 steps to actually solve. On the current problem, which I will not type out because it would take forever to get the notation and parentheses and exponents correct, I have hit a bit of a wall on simplifying. Note: I can still find the solution but it would be easier if I can implement a different strategy. My question is this: I know logarithms of like bases can be combined via VERY basic log rules -aka --- log(x) + log(x) = log(x^2) = 2 log(x) My student knows this as well. However, my question is this: Is there any relationship between more complicated expressions such as: log(x)*log(x) + log(x) which does NOT equal 2*(log(x))^2 If so, can this be extended to even more complicated expressions such as: log(A)*log(B) + log(C) where A, B, and C are all linear or exponential expressions involving the independent variable x. Note- I do not care how ugly the expression gets- I rather like ugly expressions. However, my student's knowledge is limited to Calc 2, mine is limited to ODE Thanks!
How can an ugly expression be simple?

and if any of A, B, or C are simply exponential expressions, what do you get when you take the log of an exponential expression?

 2010-07-08, 16:57 #3 Unregistered   E6B16 Posts Exponential expressions including x... For example: log(x^2-3)*log(x^3 -1) + log(2x+5)
 2010-07-08, 17:09 #4 Orgasmic Troll Cranksta Rap Ayatollah     Jul 2003 641 Posts Those are not exponential expressions. Those are polynomials. I fear for your students.
 2010-07-08, 23:57 #5 Primeinator     "Kyle" Feb 2005 Somewhere near M50..sshh! 11011111102 Posts I had a similar question myself. The only rule I know of is log (a) + log (b) = log (ab) or log (a/b) for subtraction. Orgasmic Troll, I think he/she means logarithmic arguments containing exponential terms -ex binomials or other forms of polynomials. Last fiddled with by Primeinator on 2010-07-08 at 23:59
2010-07-09, 12:18   #6
Mini-Geek
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

17·251 Posts

Quote:
 Originally Posted by Unregistered log(x)*log(x) + log(x) which does NOT equal 2*(log(x))^2 If so, can this be extended to even more complicated expressions such as: log(A)*log(B) + log(C) where A, B, and C are all linear or exponential expressions involving the independent variable x. Note- I do not care how ugly the expression gets- I rather like ugly expressions. However, my student's knowledge is limited to Calc 2, mine is limited to ODE Thanks!
Keeping in mind that x*log(y)=log(y^x) and log(x)+log(y)=log(xy):
$\log A *\log B + \log C =$
$\log{(B^{\log A})} + \log C =$
$\log(C*B^{\log A})$
Which of course is the same as $\log(C*A^{\log(B)})$ since multiplication is commutative (log(A)*log(B)=log(A^log(B))).
if x=A=B=C, (meaning the original equation was $\log x *\log x + \log x$) then that expression is equal to:
$\log(x*x^{\log x}) =$
$\log(x^{\log(x)+1})$
Hey, you said you didn't care if it gets ugly.

Last fiddled with by Mini-Geek on 2010-07-09 at 12:38

2010-07-09, 13:42   #7
R.D. Silverman

Nov 2003

11101001001002 Posts

Quote:
 Originally Posted by Orgasmic Troll Those are not exponential expressions. Those are polynomials. I fear for your students.
As do I. He/she seems incompetent.

 2010-12-05, 11:26 #8 lorgix     Sep 2010 Scandinavia 3×5×41 Posts People come here for help. They obviously know that they could be more competent.
2010-12-05, 11:45   #9
xilman
Bamboozled!

"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across

2·3·1,753 Posts

Quote:
 Originally Posted by R.D. Silverman As do I. He/she seems incompetent.
Quote:
 Originally Posted by lorgix People come here for help. They obviously know that they could be more competent.
[pontification]
When faced with these two statements, I must confess that I prefer the second. As I see it, everyone is incompetent in the sense that no-one knows everything and and can do everything in a particular field. To that extent, Bob's statement is correct but (largely) vacuous.

On the premise that everyone can improve their level of competence if they wish to do so, lorgix's statement appears to me to be much more positive.

I am concerned, as is Bob, that all too often some people are attempting to work at a level above the required degree of competence. Such people, if they are willing to take action, should be encouraged to learn and those capable of teaching them should attempt to do so. Ignorance is a curable condition.
[/pontification]

Paul

 2010-12-05, 11:55 #10 cmd     "(^r'°:.:)^n;e'e" Nov 2008 ;t:.:;^ 17478 Posts every ignorant can learn what it ignores, then every jurisdiction should learn to smile its jurisdiction
 2010-12-05, 23:05 #11 davar55     May 2004 New York City 10000100010002 Posts Ignorance is curable Tho some think unendurable Before one claims that he knows more He oughta know just what's in store

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