An algebraic quandry

2010-07-08, 15:54

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!

Orgasmic Troll · 2010-07-08, 16:53

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

Exponential expressions including x... For example:

log(x^2-3)*log(x^3 -1) + log(2x+5)

Orgasmic Troll · 2010-07-08, 17:09

Those are not exponential expressions. Those are polynomials. I fear for your students.

Primeinator · 2010-07-08, 23:57

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.

Mini-Geek · 2010-07-09, 12:18

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.

R.D. Silverman · 2010-07-09, 13:42

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.

lorgix · 2010-12-05, 11:26

People come here for help. They obviously know that they could be more competent.

xilman · 2010-12-05, 11:45

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

cmd · 2010-12-05, 11:55

every ignorant can learn what it ignores,
then every jurisdiction should learn to smile
its jurisdiction

davar55 · 2010-12-05, 23:05

Ignorance is curable
Tho some think unendurable
Before one claims that he knows more
He oughta know just what's in store

2010-07-08, 15:54	#1
Unregistered 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 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, 23:57	#5
Primeinator "Kyle" Feb 2005 Somewhere near M52.. 915₁₀ 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

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2010-07-08, 16:57	#3
Unregistered 2⁴×3×7² 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 1010000001₂ Posts	Those are not exponential expressions. Those are polynomials. I fear for your students.

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, 11:55	#10
cmd "(^r'°:.:)^n;e'e" Nov 2008 ;t:.:;^ 3³×37 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 2³·23² Posts	Ignorance is curable Tho some think unendurable Before one claims that he knows more He oughta know just what's in store