Range of natural log, some trigonometry

[Kawakaze]

Homework Statement

(1)
Real variables x and y are related by the equation
3ln(y +4) = 2ln(x +2) − 2ln(x +9)+3ln(x^2 − 1).

Determine the range of values of x and y for which the expressions on each side of this equation are defined.

Find y explicitly as a function of x, that is, express the equation in the form y = f(x), simplifying your answer as far as possible.

(2)
Show that the expressions √3cos(5t + π/6) and (√3/2)(√3cos(5t) − sin(5t)) are equivalent.

Homework Equations

None

The Attempt at a Solution

(1)
I assume y must be > -4 to make the ln of a real number. As for x I have no idea which one to use, do I have to expand them somehow? Or is it simply x^2 > 1 and x > -9 and x > -2 leaving us with the lowest x > -9

(2)
Uhm... no idea really, please give me a clue. Something tells me its substituting identities and simplification.

 

[tiny-tim]

Hi Kawakaze!

(try using the X² icon just above the Reply box )

(1) I assume y must be > -4 to make the ln of a real number. As for x … is it simply x^2 > 1 and x > -9 and x > -2

yes :smil:

leaving us with the lowest x > -9

nooooooo …

(2) Uhm... no idea really, please give me a clue. Something tells me its substituting identities and simplification.

hint: use standard trigonometric identities (and sinπ/6 = sin30° = 1/2)

 

[Hurkyl]

Or is it simply x^2 > 1 and x > -9 and x > -2

Right. To be well-defined, the argument to each logarithm must be in the domain of the logarithm. If anyone of them aren't, then the entire expression is undefined.

leaving us with the lowest x > -9

How exactly did you solve the system of three inequalities?

 

[Kawakaze]

Thanks for the tips, I solved the trig question. I am making some progress with the others, hence the edit. I doubt I am out of the woods yet though =)

 

[Kawakaze]

Hi again,

I can't do the inequality question to save my life. I also can't find how to do it on google. My best guess would be to equate it all to zero. giving -

x+2 - x+9 +x²-1 = 0

Which cancels to

x² + 10 = 0

Giving x > \sqrt{10}

I then have to rewrite y as a function of x. Would this be cube root both sides and rearrange? to give

f(x) = \stackrel{x^2 -1}{-7} +4

Which looks incorrect =(

 

[tiny-tim]

Hi Kawakaze!

I don't understand. Are you talking about …

(1)
Real variables x and y are related by the equation
3ln(y +4) = 2ln(x +2) − 2ln(x +9)+3ln(x^2 − 1).

If so, you can't add them, you have to multiply them, don't you?

 

[Hurkyl]

I can't do the inequality question to save my life.

Doesn't your textbook talk about solving inequalities? If not, drawing pictures really does help -- in fact, the solution method I thought would be standard is essentially just a picture of the number line with the solution sets drawn on it, and enough annotation to identify the edges of the solution sets (and if the edges are included).

 

[Kawakaze]

The examples in the textbook are all as easy as hell, only one inquality to solve. Then they give a test and there are 3 in one equation.

Theres not one number line in my book. The examples it gives are simple ones like range of ln(x+3), so x must be more than -3. Theres not one example with more than one inequality and not one example with a variable to a power. A total crock really. Did I mention the book also arrived a month late and this assessment is due in a week. =) They won't give an extension either.

I have got the following out x > -2, x > -9, and x² > 1

Now how to combine these, i have no idea. I would have said the range was x > -9 as this also includes the other inequalities.

 

[tiny-tim]

I have got the following out x > -2, x > -9, and x² > 1

Now how to combine these, i have no idea. I would have said the range was x > -9 as this also includes the other inequalities.

What about x = 0 ?

 

[Kawakaze]

What about it? :)

Oh wait, 0² is not greater than 1. Its x > 1 isn't it? *facepalm*

So how would this apply to arranging this as a f(x) expression? From what little I have read, solving the range of y and x should als answer this?

 

[tiny-tim]

Determine the range of values of x and y for which the expressions on each side of this equation are defined.

So how would this apply to arranging this as a f(x) expression? From what little I have read, solving the range of y and x should als answer this?

The question doesn't ask for an "f(x) expression" …

just write something like "a < x ≤ b and c < x".

 

[Kawakaze]

It does :P

Find y explicitly as a function of x, that is, express the equation in the form y = f(x), simplifying your answer as far as possible.

I was thinking of taking the cube root of both sides, but as my garbled equation above shows, that probably isn't the way to go.

Thanks for your help btw =)

 

[tiny-tim]

oh I see.

ok, simplify it as far as you can first, and leave worrying about the range until later

(it's logs, so take e-to-the of both sides)

 

[Kawakaze]

I took the exp of both sides to get

(y + 4)³ = (x + 2)² - (x + 9)² + (x² - 1)³

I then took the cube root of this to get

y + 4 = \stackrel{x^2 - 1}{-7}

So f(x) = \stackrel{x^2 - 1}{-7} -4

But this doesn't look tidy enough to be correct

 

[tiny-tim]

(y + 4)³ = (x + 2)² - (x + 9)² + (x² - 1)³

No!

You're not a natural logger, are you?

(y + 4)³ = (x + 2)²(x² - 1)³/(x + 9)² …

do you see why?​

ok, now simplify that some more!

 

[Kawakaze]

Hehe until the other day I had never really used them at all. I do not fully understand how you arrived at that. I thought multiplying you added the logs, to divide you subtract the logs. But I took the exp function, so the logs are gone, I thought i could handle the numbers as normal. Is what you did essentially go back a step, to what the equation was before the natural logs were taken? If so why would you use the exp function?

 

[tiny-tim]

ok, let's go over it carefully …

you had (ignoring the factors) lnA = lnB + lnC - lnD …

exp both sides and you get e^lnA = e^{lnB + lnC - lnD} …

now the RHS is:

e^lnB times e^lnC times e^-lnD

which you can now directly translate (btw, to answer your question, this is why we use the exp) as:

B times C times 1/D

ok, now try writing that out again, but with the factors (2 and 3) included this time

 

[Kawakaze]

Eureka, I get it! Sneaky textbooks.

(y + 4)³ = ((x + 2)² x (x² - 1)³)/(x + 9)²

Ha simplify this

y = cube root ( ((x + 2)² x (x² - 1)³)/(x + 9)² ) + 4

Sorry, latex is working against me :)

 

[tiny-tim]

Eureka, I get it! Sneaky textbooks.

(y + 4)³ = ((x + 2)² x (x² - 1)³)/(x + 9)²

Yup!

Ha simplify this

y = cube root ( ((x + 2)² x (x² - 1)³)/(x + 9)² ) + 4

Sorry, latex is working against me :)

(Actually, I prefer not to use LaTeX)

No, it's minus 4, isn't it?

And you can still simplify that fraction.

 

[Kawakaze]

Yes of course its minus 4, sorry I am a bit fried from the panic study.

y = cube root ( ((x + 2)2 x (x2 - 1)3)/(x + 9)2 ) + 4

hmmm

f(x) = ((x² - 1)/(x+2)(x+9)⁵) - 4

 

[tiny-tim]

y = cube root ( ((x + 2)2 x (x2 - 1)3)/(x + 9)2 ) + 4

(still minus, btw )

come on, think! …

what is ( (x + 2)²(x² - 1)³)(x + 9)^-2)^1/3 ??

 

[Kawakaze]

How about

(x + 2)^2/3(x² - 1)(x + 9)^-2/3 - 4

The first term is easy enough, I assume a power of ^3/3 cancels to a 1, and the power in the denominator is ^-2 x ^1/3 giving ^-2/3

 

[tiny-tim]

Hi Kawakaze!

(just got up :zzz: …)

How about

(x + 2)^2/3(x² - 1)(x + 9)^-2/3 - 4

The first term is easy enough, I assume a power of ^3/3 cancels to a 1, and the power in the denominator is ^-2 x ^1/3 giving ^-2/3

Yes, that's exactly correct!

Though personally I'd write it …

(x² - 1)(x + 2)^2/3(x + 9)^-2/3 - 4​

… because I think it looks neater! ​

 

[Kawakaze]

Hi Kawakaze!

(just got up :zzz: …)

Good morning Tim, I am also just awake, I was up till 4 am on the rest of the assessment!

Thanks so much for your help, it makes a lot more sense now :)

 

[mugaliens]

Hi Kawakaze!

(just got up :zzz: …)


Yes, that's exactly correct!

Though personally I'd write it …

(x² - 1)(x + 2)^2/3(x + 9)^-2/3 - 4​

… because I think it looks neater! ​

So this describes... what?

Please assuage my suspicions...