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Range of natural log, some trigonometry

  • Thread starter Kawakaze
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  • #1
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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.
 

Answers and Replies

  • #2
tiny-tim
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Hi Kawakaze! :smile:

(try using the X2 icon just above the Reply box :wink:)
(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 … :redface:
(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) :wink:
 
  • #3
Hurkyl
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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 any one 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?
 
  • #4
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Thanks for the tips, I solved the trig question. Im making some progress with the others, hence the edit. I doubt im out of the woods yet though =)
 
Last edited:
  • #5
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Hi again,

I cant do the inequality question to save my life. I also cant 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 > [tex]\sqrt{10}[/tex]

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

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

Which looks incorrect =(
 
  • #6
tiny-tim
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Hi Kawakaze! :smile:

I don't understand. :confused: 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? :redface:
 
  • #7
Hurkyl
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I cant 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).
 
  • #8
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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 wont 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.
 
  • #9
tiny-tim
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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 ? :redface:
 
  • #10
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What about it? :)

Oh wait, 02 is not greater than 1. Its x > 1 isnt 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?
 
  • #11
tiny-tim
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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". :smile:
 
  • #12
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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 isnt the way to go.

Thanks for your help btw =)
 
  • #13
tiny-tim
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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)
 
  • #14
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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 = [tex]\stackrel{x^2 - 1}{-7}[/tex]

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

But this doesn't look tidy enough to be correct
 
  • #15
tiny-tim
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(y + 4)3 = (x + 2)2 - (x + 9)2 + (x2 - 1)3
No!!!

You're not a natural logger, are you? :redface:

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

do you see why?

ok, now simplify that some more! :smile:
 
  • #16
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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?
 
  • #17
tiny-tim
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ok, let's go over it carefully :smile:

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

exp both sides and you get elnA = elnB + lnC - lnD

now the RHS is:

elnB times elnC 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 :wink:

ok, now try writing that out again, but with the factors (2 and 3) included this time :smile:
 
  • #18
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Eureka, I get it! Sneaky text books.

(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 :)
 
  • #19
tiny-tim
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Eureka, I get it! Sneaky text books.

(y + 4)3 = ((x + 2)2 x (x2 - 1)3)/(x + 9)2
Yup! :biggrin:
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. :wink:
 
  • #20
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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
 
  • #21
tiny-tim
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y = cube root ( ((x + 2)2 x (x2 - 1)3)/(x + 9)2 ) + 4
(still minus, btw :wink:)

come on, think! …

what is ( (x + 2)2(x2 - 1)3)(x + 9)-2)1/3 ?? :smile:
 
  • #22
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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
 
  • #23
tiny-tim
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Hi Kawakaze! :smile:

(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! :smile:

Though personally I'd write it …
(x2 - 1)(x + 2)2/3(x + 9)-2/3 - 4​
… because I think it looks neater! o:)
 
  • #24
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Hi Kawakaze! :smile:

(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 :)
 
  • #25
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Hi Kawakaze! :smile:

(just got up :zzz: …)


Yes, that's exactly correct! :smile:

Though personally I'd write it …
(x2 - 1)(x + 2)2/3(x + 9)-2/3 - 4​
… because I think it looks neater! o:)
So this describes... what?

Please assuage my suspicions...
 

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