Express y as a function of x. lny=lnx+ln(x+17)+lnC

Question

tkhunny · Answer

Can you use your regular logarithm rules?  ln(a) + ln(b) = ln(ab) -- for example?

e.mccormick · Answer

@tkhunny That won't simplify ln(x+17).  Use e.

anonymous · Answer

Use e how?

e.mccormick · Answer

$$e^{ln(a)}=a$$Raise everything to the power of e.

tkhunny · Answer

What?  Yes it will.

$\ln(x) + \ln(x+17) = ln(x\cdot(x+17))$

Why do you think it won't do that?

rajee_sam · Answer

y= x.(x+17)C

e.mccormick · Answer

@tkhunny That is not y as a function of x.  What I meant was it won't break it out of the parentheses, which is a common mistake people make when trying to change things like this.

anonymous · Answer

So I did that and now I have $$y=x+(x-17)+C$$

e.mccormick · Answer

Ah, but what you did (@tkhunny) will make the answer cleaner.  It takes more than just the regular logarithm rules, it also takes e.  Depending on if they want all logs gone or not, you can take the entire thing to e and be done, or do the regular logarithm rules then takre it to e.

tkhunny · Answer

What difference does it make whether you do it as a logarithm function or as an exponent function.  It is the same thing.

Logarithms and Exponents turn addition into multiplication.  change those addition signs to multiplication signs and you'll be there.

e.mccormick · Answer

Because that is lny as a function of x, not y. That is all the difference is. http://www.algebra.com/algebra/homework/logarithm/logarithm.faq.question.573964.html

rajee_sam · Answer

no they are asking you to write y as a function of x not lny

e.mccormick · Answer

What tkhunny is starting with is perfectly correct.  It is just incomplete.  I was wrong in saying it is incorrect.  It is just incomplete.
$$lny=lnx+ln(x+17)+lnC
\implies\
lny=ln(x⋅(x+17) C)
\implies\
e^{lny}=e^{ln(x(x+17) C)}
\implies\
y= x(x+17) C
\implies\
y= Cx^2+17Cx$$

tkhunny · Answer

It's called a hint.

BTW, why do we have ln(y) = ln(x) + ln(C)?  If we are solving a simple differential equation, we would normally get ln(y) = ln(x) + C.  Just a little different presentation.  Of course, whether it is ln(C) or just C, it is just as arbitrary.  Just thought I'd ask.

e.mccormick · Answer

@tkhunny all right. all right.  I take my lumps for that one.  I jumped the gun.  As for the C, I don't know why they decided to add it like that.  I was used to the ones where they do the +C, then doing all the LN manipulations do not matter because it prevents you from breaking it out of the exponent of e.

tkhunny · Answer

@e.mccormick  No worries.  I was just picking on you.  You proved yourself in possession of sufficient fortitude AND sufficient mathematics.  Good work.

e.mccormick · Answer

If it had been: $lny=lnx+ln(x+17)+C$ then it would be: $y=e^{lnx+ln(x+17)+C}$, which is what I have seen a few times.  No point in log manipulations there.

e.mccormick · Answer

@alienshe Do you get what we have been talking about here?  Why this happens?

tkhunny · Answer

I would have to disagree on what I normally tell students being introduced to logarithms:  "Logarithms ARE Exponents"  There just isn't a difference.  :-)

e.mccormick · Answer

@tkhunny In the same sense that there is no subtraction.  I have a Linear Algebra prof that insists there is no such thing as subtraction.  Just shorthand notation of adding a negative.

tkhunny · Answer

We all have little hang-ups.  One of mine is the absence of a difference between logarithms and exponents.

e.mccormick · Answer

I do radio stuff.  I am kind of used to seeing log shorthand.  It is used to simplify things there.