How to simplify ln ( (x^3)((1-x)^3/2) / ((x+1)^1/5) )

- anonymous

How to simplify ln ( (x^3)((1-x)^3/2) / ((x+1)^1/5) )

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- anonymous

this was originally: 3lnx + 1/5ln(x+1) + 3/2ln(1-x) and i simplified to what I types above

- anonymous

Is this what you mean?
\(\dfrac{x^3(1-x)^3}{2}\div \dfrac{(x+1)^2}{5}\)

- anonymous

\[\ln (\frac{ x ^{3} \times (1-x)^{3/2} }{ (x+1)^{1/5} } )\]

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## More answers

- DebbieG

Started with:
\[3\ln x + \dfrac{1}{5}\ln(x+1) + \dfrac{3}{2}\ln(1-x) \]

- DebbieG

All the log expressions are added in the original form, right? so there shouldn't be a quotient involved.....

- anonymous

well from the added expressions, I simplified it to what I posted above (the fraction)

- 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

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

the quotient was from the first variable - last variable
according to the laws, log-log = log/log

- 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

oh damn, I wrote it incorrectly, it was supposed to be the first variable + second variable - third variable
oops!

- DebbieG

There is no difference of log expressions there.

- DebbieG

ah-HAH, well then that will make a big difference! (pun intended! :) lol

- DebbieG

so it's
\[3\ln x + \dfrac{1}{5}\ln(x+1) - \dfrac{3}{2}\ln(1-x)\]
right?

- 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

or is it\[3\ln x - \dfrac{1}{5}\ln(x+1) + \dfrac{3}{2}\ln(1-x)\]

- 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

\[\ln (\frac{ x ^{3} \times (1-x)^{3/2} }{ (x+1)^{1/5} } )\]
would be the result after simplifying
\[3\ln x - \dfrac{1}{5}\ln(x+1) + \dfrac{3}{2}\ln(1-x)\]

- anonymous

okay, here is the correct equation: \[3lnx - 1/5\ln(x+1) + 3/2\ln(1-x)\]

- DebbieG

(but to answer your other question, no, that mess inside the ln( ) can't be simplified further.)

- DebbieG

OK, then you're golden! :)

- 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

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

E.g.:\[\Large (1-x)^{3/2} =\sqrt{(1-x)^3}\]

- anonymous

so it would be: \[\ln (\sqrt[3]{x} \times (1-x)^{3/2} \div \sqrt[5]{x+1})\]

- anonymous

of course, the 1-x would be as you mentionned above as well

- 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

it's not root3 and root(x+1)^5 right?

- 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

um, so \[x^3 = 3\sqrt{x}\] ?

- 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

and \[(1-x)^\frac{ 3 }{ 2 } = \sqrt{(1-x)^3 ?

- anonymous

wait nvm lol

- DebbieG

\[\Large (1-x)^\frac{ 3 }{ 2 } = \sqrt{(1-x)^3} \]

- anonymous

ya, I did that for the (1-x) part, but for the x^3, does it equal to root3 ?

- anonymous

NVM THAT IS WRONG LOL. IM STUPID

- 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

thanks! I will remember!!!

- DebbieG

Good. then my work here is done... lol. :)

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