Another chain rule with logs

[Dorian Gray]

Greetings fellow Mathematicians,

I was wondering if someone could please take a look at my work. The problem was ln [(2x+1)^3/(3x-1)^4].


My answer key says [6/(2x+1)]-[12/(3x-1], which is different from what I got.

Thanks for any help, comments, suggestions, etc.

 

[pka]

Greetings fellow Mathematicians,

I was wondering if someone could please take a look at my work. The problem was ln [(2x+1)^3/(3x-1)^4].

Simplify first:

\(\displaystyle \ln \left[ {\dfrac{{(2x + 1)^3 }}{{(3x - 1)^4 }}} \right] = 3\ln (2x + 1) - 4\ln (3x - 1)\)

 

[Deleted member 4993]

Greetings fellow Mathematicians,

I was wondering if someone could please take a look at my work. The problem was ln [(2x+1)^3/(3x-1)^4].
View attachment 1711

My answer key says [6/(2x+1)]-[12/(3x-1], which is different from what I got.

Thanks for any help, comments, suggestions, etc.

First use these facts:

ln(U/V) = ln(U) - ln(V)

ln(U^m) = m * ln(U)

Now your answer will be much simpler - like the key suggests.....

 

[Dorian Gray]

Thank you both

Once again, thank you both Pka and and Subhotosh for your input. I can see where that makes things much, much better to work with. I will most certainly redo the problem that way. Out of curiosity, are they still the same answer?

 

[Dorian Gray]

 

[srmichael]

View attachment 1712

Yes, your ugly answer and \(\displaystyle \displaystyle\frac{6}{2x+1}-\frac{12}{3x-1}\) are the same answer.

On your post above, there is no need to do the product rule on each of these terms. The 3 and 4 in front of the natural log terms are just constants so:

\(\displaystyle \displaystyle y=3\ln(2x+1)-4\ln(3x-1)\)

\(\displaystyle \displaystyle y'=3\cdot\frac{2}{2x+1}-4\cdot\frac{3}{3x-1}\)

\(\displaystyle \displaystyle y'=\frac{6}{2x+1}-\frac{12}{3x-1}\)

Make sense?

 

[Dorian Gray]

Thank you SirMichael

Thank you SirMichael. I appreciate seeing your work. Although, when I saw it, I think I did my 5th face palm of the day