Question

Express as one logarithm
\[3ln(\frac{y^2}{x})-2ln(y)+\frac{1}{3}ln(\frac{x^6y^3}{2})\]

Answer 1

I got\[ \ln(\frac{y^2\sqrt[3]{2}}{x^5})\]

Answer 2

\[its \ln(\frac{y^{5}}{x (2^{1/3})})\]

Answer 3

How so?

Answer 4

gimme me minute!!

Answer 5

Okay :)

Answer 6

\[\ln(\frac{y^{6}}{x^{3}})-\ln (y^{2})+\ln(\frac{x^{2}y}{2^1/3})\]

Answer 7

\[6\ln(\frac{y^{6}*x^{2}*y}{x^{3}*y^{2}*2^{1/3}})\]

Answer 8

please ignore the 6..

Answer 9

finally u arrive at the given answer...

Answer 10

your answer or mine?

Answer 11

mine....

Answer 12

cancel the terms....

Answer 13

But...ln(y^6/x^3)=ln(y^6x^-3)...

Answer 14

yeah so??

Answer 15

\[3\ln(\frac{y^2}{x})-2\ln(y)+\frac{1}{3}\ln(\frac{x^6y^3}{2}) \]

Recall \[r \ln(u)=\ln(u^r)\]
So first I will do:
\[\ln((\frac{y^2}{x})^3)-\ln(y^2)+\ln((\frac{x^6y^3}{2})^\frac{1}{3})\]
Now using law of exponents we have
\[\ln(\frac{y^6}{x^3})-\ln(y^2)+\ln(\frac{x^2y}{2^\frac{1}{3}})\]
So Also recall ln(a)-ln(b)=ln(ab)
Giving us:
\[\ln(\frac{y^6}{x^3} \div y^2)+\ln(\frac{x^2y}{\sqrt[3]{2}})\]
Now recall that a/b divided by c/d =a/b * d/c
\[\ln(\frac{y^6}{x^3} \cdot \frac{1}{y^2})+\ln(\frac{x^2 y}{\sqrt[3]{2}})\]
Then blah blah
I don't now if you guys have reached a result but if you haven't It should be pretty easy to continue from here

Answer 16

oops type-o

Answer 17

ln(a)-ln(b)=ln(a/b)

Answer 18

Is that a 3 in the denominator of your first ln?

Answer 19

Are you talking the cube or the cube root?

Answer 20

The answer I got is all the way to the top..

Answer 21

Let me know what you don't understand

Answer 22

why my answer is wrong...

Answer 23

I don't know where you went wrong unless you show me your work

Answer 24

This is what I did
\[\ln(\frac{y^6}{x^3})-\ln(y^2)+\ln(\frac{x^2y}{\sqrt[3]{2}})\]\[\ln(y^6x^{-3})-\ln(\frac{x^2y^3}{\sqrt[3]{2}})\]\[\ln(y^6x^{-3})-\ln(\frac{x^2y^3}{\sqrt[3]{2}})\]\[\ln(\frac{y^6x^{-3}\sqrt[3]{2}}{x^2y^3})\]\[\ln(\frac{y^2\sqrt[3]{2}}{x^5})\]

Answer 25

So but you are saying 6-3=2 ?
let me see if i see any other mistakes

Answer 26

Okay..

Answer 27

and how did you get a minus in front of the last natural log on your 2nd line?

Answer 28

It was always there, I brought the two second terms together using the product law of logarithms

Answer 29

in how did you get a 3 as exponent for that one value of y?

Answer 30

ok no no. you did something wrong there then...let me show ya

Answer 31

\[\ln(\frac{y^6}{x^3})-\ln(y^2)+\ln(\frac{x^2y}{2^\frac{1}{3}}) \]
We agreed to this point right @purplec16 ?

Answer 32

Yes...

Answer 33

I can write it like this this because a+b=b+a
addition is commutative
\[\ln(\frac{y^6}{x^3})+\ln(\frac{x^2y}{\sqrt[3]{2}})-\ln(y^2)\]

Answer 34

yup

Answer 35

Now you worked with the second two terms first right?

Answer 36

So that is what I'm gonna do in this next step

Answer 37

Yes

Answer 38

\[\ln(\frac{y^6}{x^3})+\ln(\frac{x^2 y}{\sqrt[3]{2}} \div y^2)\]
by that one property of log : \[\ln(a \div b) \text{ or } \ln(\frac{a}{b})=\ln(a)-\ln(b)\]

Answer 39

\[\ln(\frac{y^6}{x^3})+\ln(\frac{x^2 y}{\sqrt[3]{2}} \cdot \frac{1}{y^2})\]

Answer 40

do you see what to do from here?

Answer 41

I guess... am so tired.. I will try to figure it out in the mornign because I am not seeing it but thanks for all your help!

Answer 42

which one of the steps I did didn't make sense?

Answer 43

the 2nd term of your last equation

Answer 44

do you know that a/b divided by c/d=a/b * d/c ?

Answer 45

yes