anonymous
  • anonymous
How to simplify ln ( (x^3)((1-x)^3/2) / ((x+1)^1/5) )
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
this was originally: 3lnx + 1/5ln(x+1) + 3/2ln(1-x) and i simplified to what I types above
anonymous
  • anonymous
Is this what you mean? \(\dfrac{x^3(1-x)^3}{2}\div \dfrac{(x+1)^2}{5}\)
anonymous
  • anonymous
\[\ln (\frac{ x ^{3} \times (1-x)^{3/2} }{ (x+1)^{1/5} } )\]

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More answers

DebbieG
  • DebbieG
Started with: \[3\ln x + \dfrac{1}{5}\ln(x+1) + \dfrac{3}{2}\ln(1-x) \]
DebbieG
  • DebbieG
All the log expressions are added in the original form, right? so there shouldn't be a quotient involved.....
anonymous
  • anonymous
well from the added expressions, I simplified it to what I posted above (the fraction)
DebbieG
  • DebbieG
It looks like you handled all the coefficients correctly by changing them to exponents inside the ln function.... but I don't see where you got a quotient. :)
DebbieG
  • DebbieG
ln(M) + ln(N) = ln(MN) ln(M) - ln(N) = ln(M/N) You don't have a difference of logs anywhere (as I read your original expression), so you won't get a quotient inside the log, just product.
anonymous
  • anonymous
the quotient was from the first variable - last variable according to the laws, log-log = log/log
DebbieG
  • DebbieG
Is this correct, for what you started with? \[3\ln x + \dfrac{1}{5}\ln(x+1) + \dfrac{3}{2}\ln(1-x)\]
anonymous
  • anonymous
oh damn, I wrote it incorrectly, it was supposed to be the first variable + second variable - third variable oops!
DebbieG
  • DebbieG
There is no difference of log expressions there.
DebbieG
  • DebbieG
ah-HAH, well then that will make a big difference! (pun intended! :) lol
DebbieG
  • DebbieG
so it's \[3\ln x + \dfrac{1}{5}\ln(x+1) - \dfrac{3}{2}\ln(1-x)\] right?
anonymous
  • anonymous
yes, that is correct :P sorry! I was convinced I typed it correctly. Anywhoo, would my ln(fraction) be correct? or is there another way to simplify it even more?
DebbieG
  • DebbieG
or is it\[3\ln x - \dfrac{1}{5}\ln(x+1) + \dfrac{3}{2}\ln(1-x)\]
DebbieG
  • DebbieG
If the 3rd term is the one that is subtracted,t hen why is it in the numerator? and the 2nd term in your den'r?
DebbieG
  • DebbieG
\[\ln (\frac{ x ^{3} \times (1-x)^{3/2} }{ (x+1)^{1/5} } )\] would be the result after simplifying \[3\ln x - \dfrac{1}{5}\ln(x+1) + \dfrac{3}{2}\ln(1-x)\]
anonymous
  • anonymous
okay, here is the correct equation: \[3lnx - 1/5\ln(x+1) + 3/2\ln(1-x)\]
DebbieG
  • DebbieG
(but to answer your other question, no, that mess inside the ln( ) can't be simplified further.)
DebbieG
  • DebbieG
OK, then you're golden! :)
anonymous
  • anonymous
thanks! Are you sure I cannot simplify more? I feel like there is a clearer way to write the exponents through roots, but I don't know how!
DebbieG
  • DebbieG
Oh, well, you could put into radical form. That's not really "simplifying", just different notation. E.g in the numerator you'd have a cube under a square root; in the den'r you'd put that whole thing under a 5th root sign. :)
DebbieG
  • DebbieG
E.g.:\[\Large (1-x)^{3/2} =\sqrt{(1-x)^3}\]
anonymous
  • anonymous
so it would be: \[\ln (\sqrt[3]{x} \times (1-x)^{3/2} \div \sqrt[5]{x+1})\]
anonymous
  • anonymous
of course, the 1-x would be as you mentionned above as well
DebbieG
  • DebbieG
Yup, that's equivalent, you can certainly use the radicals if you prefer. (unless your teacher gave you specific instructions about the form of the answer). :)
anonymous
  • anonymous
it's not root3 and root(x+1)^5 right?
DebbieG
  • DebbieG
Numerator of a rational exponent is always the power, den'r is always the root. :) so 3/2 is the third power, square root
anonymous
  • anonymous
um, so \[x^3 = 3\sqrt{x}\] ?
DebbieG
  • DebbieG
noooo..... the exponent "3" is really 3/1 if you want to think of it as a rational exponent. So the 3 is in the num'r, hence it's a power! No root!
anonymous
  • anonymous
and \[(1-x)^\frac{ 3 }{ 2 } = \sqrt{(1-x)^3 ?
anonymous
  • anonymous
wait nvm lol
DebbieG
  • DebbieG
\[\Large (1-x)^\frac{ 3 }{ 2 } = \sqrt{(1-x)^3} \]
anonymous
  • anonymous
ya, I did that for the (1-x) part, but for the x^3, does it equal to root3 ?
anonymous
  • anonymous
NVM THAT IS WRONG LOL. IM STUPID
DebbieG
  • DebbieG
lol... no, not stupid... just remember: num'r is the power! den'r is the root! \[\Large x^\frac{ 1 }{ 3 }=\sqrt[3]{x}\]
anonymous
  • anonymous
thanks! I will remember!!!
DebbieG
  • DebbieG
Good. then my work here is done... lol. :)

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