Question

i got stuck here with my logs

Answer 1

\[-2\ln(v-1)+3\ln(v-2)=-\ln|(x)|+c\]

Answer 2

then i made some sub with v= y/x which i got this
\[3\ln(\frac{ y }{ x }-2)-2\ln(\frac{ y }{ x }-1)=-\ln|x|+c\]

Answer 3

@Zepdrix

Answer 4

Still hasn't noticed smh

Answer 5

lol he will come eventually x'D so no worries

Answer 6

Hopefully
Can't help you bc too hard for my brain

Answer 7

Oh you're trying to solve for y, yes?

Answer 8

\[\large\rm 3\ln\left[\frac{y}{x}-2\right]-2\ln\left[\frac{y}{x}-1\right]=-\ln|x|+c\]Apply your coefficient/exponent rule for logs, whatever it's called,\[\large\rm \ln\left[\left(\frac{y}{x}-2\right)^3\right]-\ln\left[\left(\frac{y}{x}-1\right)^2\right]=-\ln|x|+c\]

Answer 9

Then apply your subtraction/division rule for logs,\[\large\rm \ln\left[\frac{\left(\frac{y}{x}-2\right)^3}{\left(\frac{y}{x}-1\right)^2}\right]=-\ln|x|+c\]

Answer 10

Then exponentiate,\[\large\rm \frac{\left(\frac{y}{x}-2\right)^3}{\left(\frac{y}{x}-1\right)^2}=e^{-\ln|x|+c}\]

Answer 11

how come the right side didnt cancel with the e

Answer 12

It will, but we have to split up the exponent first.\[\large\rm \frac{\left(\frac{y}{x}-2\right)^3}{\left(\frac{y}{x}-1\right)^2}=e^{-\ln|x|}\cdot e^c\]And also deal with the negative in front,\[\large\rm \frac{\left(\frac{y}{x}-2\right)^3}{\left(\frac{y}{x}-1\right)^2}=e^{\ln|x^{-1}|}\cdot e^c\]Now they can "cancel out",\[\large\rm \frac{\left(\frac{y}{x}-2\right)^3}{\left(\frac{y}{x}-1\right)^2}=\frac{1}{x}\cdot e^c\]

Answer 13

We'll simply that last part by calling it something else,\[\large\rm \frac{\left(\frac{y}{x}-2\right)^3}{\left(\frac{y}{x}-1\right)^2}=\frac{1}{x}\cdot C\]Where C is non-zero, I suppose. Blah whatever, let's just try to get to some answer :D lol

Answer 14

Multiplying by our denominator,\[\large\rm \left(\frac{y}{x}-2\right)^3=\frac{1}{x}\cdot C\cdot \left(\frac{y}{x}-1\right)^2\]

Answer 15

To be honest, it's probably easier to do all of these previous steps leaving v as v.

Answer 16

\[\large\rm \left(v-2\right)^3=\frac{1}{x}\cdot C\cdot \left(v-1\right)^2\]

Answer 17

oh yeah thats right e.e

Answer 18

And then ummm.. hmm <.<
Do we have to expand and do all that ugly business? Hmmm

Answer 19

hmmmm it also says y(0) = 2
:3 .. do you wanna see the original problem

Answer 20

Yes pls

Answer 21

Mmmm I'm getting a little stuck ...
It doesn't seem like you can make the substitution v=y/x because v is not defined at x=0.

So y(0)=2 doesn't work for our substitution. Hmm

Answer 22

right hmmm

Answer 23

Oh... I guess this one is non-linear.
https://www.wolframalpha.com/input/?i=y%27(x)%3D(2*(2y(x)-x)%2F(x%2By(x))),+y(0)%3D2

So we're not going to be able to get a solution y(x).
We'll have to leave it as something ugly, I think :U
Hmm

Answer 24

yeh D:

Answer 25

Weird my link broke :|

Answer 26

D:

Answer 27

at a glance, that looks very do-able. so, i copied and pasted, and it blows up Wolfram.

now, isn't that interesting?!?! personally, i'd give it another go.

Answer 28

Were you able to get an answer Macha?
It's pretty straight forward from this point,\[\large\rm \left(\frac{y}{x}-2\right)^3=\frac{1}{x}\cdot C\cdot \left(\frac{y}{x}-1\right)^2\]
Just a few simple Algebra steps :)

Answer 29

yeah i got that x'D so then D:

Answer 30

Ok this next step might be a lil confusing.
Just try to think.

It would be nice if we could get the x's out of the denominators.
See how we have a bunch of y/x stuff?
Ya let's try to clean that up.

So think.... what is the largest power of x in the denominators?

Answer 31

hmmm we would have to factor it right

Answer 32

No, don't do that.

Answer 33

Just try to think ahead, what is the largest power of x in the denominators?

Answer 34

1?

Answer 35

It's 3.

Answer 36

So we'll multiply both sides by x^3 to clean things up.

Answer 37

woah how did that happen x'D

Answer 38

You have to THINK AHEAD.
If you were to expand everything out, then (y/x)^3 would be the largest term, yes?

So x^3 is the biggest degree of x in a denominator.

Answer 39

oh yeah thats right

Answer 40

\[\large\rm \color{orangered}{x^3}\cdot\left(\frac{y}{x}-2\right)^3=\color{orangered}{x^3}\cdot\frac{1}{x}\cdot C\cdot \left(\frac{y}{x}-1\right)^2\]

Answer 41

Simplify the x's on the right,\[\large\rm \color{orangered}{x^3}\cdot\left(\frac{y}{x}-2\right)^3=x^2\cdot C\cdot \left(\frac{y}{x}-1\right)^2\]

Answer 42

So ... another tricky Algebra step...
You want to bring this x into the cube.

Answer 43

You lose that 3rd power when you bring the x into the cube.
I hope the a,b,c picture helps to illustrate why.

Answer 44

Same type of thing will happen on the right side,
you're bring a squared x into a square.

Answer 45

\[\large\rm x^3\left[\left(\frac{y}{x}-2\right)\right]^3=Cx^2 \left[\left(\frac{y}{x}-1\right)\right]^2\]
\[\large\rm \left[x\left(\frac{y}{x}-2\right)\right]^3=C\left[x\left(\frac{y}{x}-1\right)\right]^2\]And then distribute the x to each term in the round brackets,\[\large\rm \left[y-2x\right]^3=C\left[y-x\right]^2\]

Answer 46

oh okay that makes sense. so then

Answer 47

When they provide us with the information y(0)=2,
it means that the point (0,2) satisfies this differential equation.

So we can plug in the point (0,2) to solve for C.

Answer 48

\[\large\rm \left[2-2\cdot0\right]^3=C\left[2-0\right]^2\]

Answer 49

\[\large\rm [2]^3=C[2]^2\]\[\large\rm 2=C\]

Answer 50

Go BACK to your general solution,
\[\large\rm \left[y-2x\right]^3=C\left[y-x\right]^2\]and plug this new found C value in,\[\large\rm \left[y-2x\right]^3=2\left[y-x\right]^2\]
Boom there it is.
You've found the solution to this differential equation which corresponds with the specific data they provided.

Answer 51

yay ;p

Answer 52

zeppers how would you write this ?

(320.7 million) (72.36)

Answer 53

@Zepdrix

Answer 54

I don't understand what you're asking.
Do you need to multiply, and then convert to scientific notation?

Answer 55

well i have to multiply it

Answer 56

(320.7 million) x (72.36)

= (320.7 x 72.36) million
= 23205.852 million

Which could be written in billions like this
== 23.205852 billion

Answer 57

would the 320.7 million be written as 320 700 000?

Answer 58

Yes :)

Answer 59

would the 72.36 be considered as a million too?

Answer 60

No, I don't think so.

Answer 61

aahh

okay so how did you get a billion?

i got this
2.3205852E10

Answer 62

E is a weird calculator notation.
It means your number is being multiplied by 10 to this E(xponent) power.

For example:
2.44E5 is really \(\large\rm 2.44\times10^5\)
2.773E3 is really \(\large\rm 2.773\times10^3\)

So your number,
2.3205852E10 is really \(\large\rm 2.3205852\times10^{10}\)

Answer 63

Try to get comfortable with your powers of 10.

10 to the 6 power is a million.
7.4E6 is really \(\large\rm 7.4\times10^6\) which is 7.4 million.

10 to the power 9 is a billion.
2.3205852E9 is really \(\large\rm 2.3205852\times10^9\) which is 2.3205852 billion.

If we multiply that value by 10 (because we have E10 not E9)
we just move the decimal place one to the right.

Answer 64

so its kidn of similar to this sheet

Answer 65

Yes :)

Answer 66

ah okay so then im assuming that 10 ^9 , 10 ^10 , 10^11 is known as a billion

Answer 67

Yes.
10^9 is billions
10^10 is ten billions
10^11 is hundred billions