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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Lasercookie on October 25, 2011, 05:35:07 pm

Title: laseredd's methods questions
Post by: Lasercookie on October 25, 2011, 05:35:07 pm: I'm having trouble with this question:

Given $3^x = 4^y = 12^z$ show that $z = \frac{xy}{x+y}$

My approach was to start off with using the log relationship and end up with equations like this:
$log_4(12^z)=y$
$log_3(12^z)=x$

$log_{12}(4^y)=z$
$log_{12}(3^x)=z$

and from that get stuff like:
$z log_{4}(12)=y$
$z log_{3}(12)=x$

From this point I then try substituting and rearranging stuff, but I haven't gotten anything close to the answer yet. :/

Any advice on how I should proceed?
Title: Re: laseredd's methods questions
Post by: aznxD on October 25, 2011, 05:51:00 pm: $3^x = 4^y = 12^z$

$3^x = 12^z$

$3 = 12^\frac{z}{x}$ (1)

$4^y =12^z$

$4 = 12^\frac{z}{y}$ (2)

Multiply equation (1) and (2).

$12 = 12^{\frac{z}{x}+\frac{z}{y}$

Equating the powers.

$1 = \frac{z}{x} + \frac{z}{y}$

$z = \frac{xy}{x+y}$
Title: Re: laseredd's methods questions
Post by: Lasercookie on October 25, 2011, 06:25:25 pm: Thanks for the response - that seemed quite simple :)

I have one question though:
Quote from: aznxD on October 25, 2011, 05:51:00 pm
$3^x = 12^z$

$3 = 12^\frac{z}{x}$ (1)
I was under the impression that to be able to do something like that, the bases had to be equal.

Isn't $3^x = 12^z$ the same as $3^x - 12^z = 0$ ?
To subtract stuff, don't the base and the exponent have to be the same?

I'm not entirely sure of what's actually going on here.
Title: Re: laseredd's methods questions
Post by: golden on October 25, 2011, 07:27:38 pm: Hey Laseredd.

Raise both sides to the power of 1/x.
So 3^(x) = 12^z.
Becomes 3^(x/x) = 12^(z/x)
= 3^1 = 3 = 12^(z/x).

3^x = 12^z
ln(3^x) = ln(12^z)
x(ln3) = ln(12^z)
ln(3) = ln(12)z/x
ln(3) = ln(12^(z/x)).
3 = 12^(z/x).

Someone please confirm the last part.

Edit: The last part was just included in case you were wondering.
Title: Re: laseredd's methods questions
Post by: Lasercookie on October 25, 2011, 08:05:08 pm: Thanks golden, makes a lot of sense (well the first part anyway, I haven't learnt about natural logarithms properly yet).
Title: Re: laseredd's methods questions
Post by: brightsky on October 25, 2011, 08:14:54 pm: @golden - what you did for the last bit is correct but not really necessary. your first explanation is fine, not so say more concise. it's just like taking the square root of both sides (given that both bases are positive), except in this case you're taking the xth root of both sides, and since both bases are already a positive constant, it's mathematically correct.
Title: Re: laseredd's methods questions
Post by: golden on October 26, 2011, 09:39:52 am: Quote from: brightsky on October 25, 2011, 08:14:54 pm
@golden - what you did for the last bit is correct but not really necessary. your first explanation is fine, not so say more concise. it's just like taking the square root of both sides (given that both bases are positive), except in this case you're taking the xth root of both sides, and since both bases are already a positive constant, it's mathematically correct.

Yeah I realised lol but he was referencing something about bases then I thought he was he was wanting to use eg. log base 10 (x) or something, so decided to appeal to him from that perspective. I just realised he was talking about bases as in base^x. LOL.