20100708, 15:54  #1 
7×281 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 4050 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! 
20100708, 16:53  #2  
Cranksta Rap Ayatollah
Jul 2003
641_{10} Posts 
Quote:
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? 

20100708, 16:57  #3 
2^{4}×3×7^{2} Posts 
Exponential expressions including x... For example:
log(x^23)*log(x^3 1) + log(2x+5) 
20100708, 17:09  #4 
Cranksta Rap Ayatollah
Jul 2003
1010000001_{2} Posts 
Those are not exponential expressions. Those are polynomials. I fear for your students.

20100708, 23:57  #5 
"Kyle"
Feb 2005
Somewhere near M52..
915_{10} 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 20100708 at 23:59 
20100709, 12:18  #6  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17×251 Posts 
Quote:
Which of course is the same as since multiplication is commutative (log(A)*log(B)=log(A^log(B))). if x=A=B=C, (meaning the original equation was ) then that expression is equal to: Hey, you said you didn't care if it gets ugly. Last fiddled with by MiniGeek on 20100709 at 12:38 

20100709, 13:42  #7 
Nov 2003
2^{2}·5·373 Posts 

20101205, 11:26  #8 
Sep 2010
Scandinavia
3×5×41 Posts 
People come here for help. They obviously know that they could be more competent.

20101205, 11:45  #9  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
2×17×313 Posts 
Quote:
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 noone 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 

20101205, 11:55  #10 
"(^r'Β°:.:)^n;e'e"
Nov 2008
;t:.:;^
3^{3}×37 Posts 
every ignorant can learn what it ignores,
then every jurisdiction should learn to smile its jurisdiction 
20101205, 23:05  #11 
May 2004
New York City
2^{3}·23^{2} 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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