Split - logarithmic differentiation

nikhtas30

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[MATH]\log{y} = \dfrac{1}{2}\log{x} + (x^2-x) + \dfrac{3}{2}\log(x+1)[/MATH]
The right way to writte it down is this:
\(\displaystyle \log{y} = \dfrac{1}{2}\log{x} +(x^2-x)loge+ \dfrac{3}{2}\log(x+1)\)

Because: \(\displaystyle lne=1\) and \(\displaystyle loge\neq 1\)
 
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Unfortunately, calculus, the context of this question, is at the lower boundary of advanced math, and ln is still commonly used there. So it's wise to state your definition before using "log" with any particular meaning, just to avoid confusion (and to avoid collecting unnecessary "corrections").

The right way to write it down is this:
\(\displaystyle \log{y} = \dfrac{1}{2}\log{x} +(x^2-x)\log e+ \dfrac{3}{2}\log(x+1)\)

Because: \(\displaystyle \ln e=1\) and \(\displaystyle \log e\neq 1\)
Actually, in calculus it's far better to use \(\ln\), or \(\log_e\) -- whatever you choose to call it! And that is what was intended.
 
The right way to writte it down is this:
\(\displaystyle \log{y} = \dfrac{1}{2}\log{x} +(x^2-x)loge+ \dfrac{3}{2}\log(x+1)\)

Because: \(\displaystyle lne=1\) and \(\displaystyle loge\neq 1\)
Excuse me, but who do you think you are to insinuate that a tutor/volunteer such as skeeter did not do it the right way?
I know that I sound pissed off, (and I am) but if you want to help out here that is great but please be considerate to the other tutors. Skeeter has become a legend over the years and when something is written by skeeter you take it as correct.
 
Excuse me, but who do you think you are to insinuate that a tutor/volunteer such as skeeter did not do it the right way?
I know that I sound pissed off, (and I am) but if you want to help out here that is great but please be considerate to the other tutors. Skeeter has become a legend over the years and when something is written by skeeter you take it as correct.
I don't doubt that he knows the answer and the proper way to do it.
He just didn't writte it down the right way.
 
I suggest posts 4 through 13 be split off as totally useless to the student asking the question.
 
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