laseredd's methods questions

[Lasercookie]

I'm having trouble with this question:

Given show that


My approach was to start off with using the log relationship and end up with equations like this:



   


and from that get stuff like:



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?

[aznxD]

      (1)



      (2)

Multiply equation (1) and (2).



Equating the powers.

[Lasercookie]

Thanks for the response - that seemed quite simple

I have one question though:

      (1)

I was under the impression that to be able to do something like that, the bases had to be equal.

Isn't the same as ?
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.

[golden]

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.

[Lasercookie]

Thanks golden, makes a lot of sense (well the first part anyway, I haven't learnt about natural logarithms properly yet).

[brightsky]

@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.

[golden]

@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.