# Range of natural log, some trigonometry

## 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.

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
Homework Helper
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
Staff Emeritus
Gold Member
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?

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:
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 > $$\sqrt{10}$$

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
Homework Helper
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
Staff Emeritus
Gold Member
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).

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.

tiny-tim
Homework Helper
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 ? 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?

tiny-tim
Homework Helper
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". 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 =)

tiny-tim
Homework Helper
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)

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
Homework Helper
(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! 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
Homework Helper
ok, let's go over it carefully 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 ok, now try writing that out again, but with the factors (2 and 3) included this time 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 :)

tiny-tim
Homework Helper
Eureka, I get it! Sneaky text books.

(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. 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
Homework Helper
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 ?? (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
Homework Helper
Hi Kawakaze! (just got up :zzz: …)

(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! 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 :)

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?