Uni Stuff > Mathematics

Is this working right/Random questions

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shinny:
I'm not sure if theres a faster proof for that rule but here's one:
Let:

Log both sides:

Chuck the power down using that other log law:

Simplify the log:

Equate the logs:

So therefore:

Oh, and if you're seriously struggling, try and derive all the log laws yourself through exponentials and it makes things alot more obvious why they're like that. My teacher for both years so far is really into proper understanding, so I've had to derive basically every formula I've used in spesh and methods so far, and yeh, it actually does help.

EDIT: Ok whoops, you mentioned you weren't too great at exponentials either. In that case, go back to the basic level and write them out as they are in basic form. i.e. 2^3=2*2*2, and especially being able to distinguish between (2^3)*(2^4) and (2^3)^4. Then experiment a bit more I guess and you'll see why things are like that =P Either that or just rote learn them and hope for the best. I find the exponential laws to be quite logical, but yeh, the log ones are kinda weird.

cara.mel:
why can you log both sides and drop both logs

thank you =)

Flaming_Arrow:


so they both have same log base

shinny:
Well for log'ing both sides, its just like with any function (such as square root, squaring etc); if you do it to both sides as a whole, then yeh, its still equal (some exceptions though I imagine). As for dropping powers...hmmm..I'll get a proof for that one, wait a sec =P

EDIT: ok mao beat me to it =T

Mao:
the definition of a logarithm is the index the base must be raised to to get to that number,

i.e. if , the base "a" must be raised to the bth power to get to c,

using this principle, if we raise the base by the power of the base:



and,

also, from the few index laws:

(this is the first log law)

the second one can be shown in a similar fashion.

the third law lies on the fact that for non-zero a. hence, the log of 1 of any non-zero base is 1.

the fourth law is evidently true:

the fifth is a combinations of the previous

the next one uses

and the last one is tricky:
, where k is a constant such that

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