Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

Started with:

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

Answer 5

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

Answer 6

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

Answer 7

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. :)

Answer 8

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.

Answer 9

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

Answer 10

Is this correct, for what you started with?
\[3\ln x + \dfrac{1}{5}\ln(x+1) + \dfrac{3}{2}\ln(1-x)\]

Answer 11

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

Answer 12

There is no difference of log expressions there.

Answer 13

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

Answer 14

so it's

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

right?

Answer 15

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?

Answer 16

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

Answer 17

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?

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

OK, then you're golden! :)

Answer 22

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!

Answer 23

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. :)

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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). :)

Answer 28

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

Answer 29

Numerator of a rational exponent is always the power, den'r is always the root. :)

so 3/2 is the third power, square root

Answer 30

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

Answer 31

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!

Answer 32

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

Answer 33

wait nvm lol

Answer 34

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

Answer 35

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

Answer 36

NVM THAT IS WRONG LOL. IM STUPID

Answer 37

lol... no, not stupid... just remember:

num'r is the power!
den'r is the root!

\[\Large x^\frac{ 1 }{ 3 }=\sqrt[3]{x}\]

Answer 38

thanks! I will remember!!!

Answer 39

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