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Discussion in 'MCAT Study Question Q&A' started by inaccensa, Aug 6, 2011.
The answer is -2
can someone please explain this
You can solve for X explicitly by taking the base-2 logarithm of both sides, but you should probably be able to realize that 1/4 is just 1/2^2 = 2^(-2) by inspection.
great thanks. i know its simple,but I couldnt understand it
so its xlog 2 = log (1/4)
x log 2 = log (-4)
just substituting the numbers, right
first of all, there is no log of a negative number. log(-4) doesnt exist. because exponent of something is always positive.
secondly, forget the log, it will just make it more confusing, there is no reason to solve for log with no other base than 10...
1/(2^2) = 2^(-2)
anytime you see X^Y = 1/Z
just solve for X^Y = Z, then put a negative on the Y.
yeah I meant in the next step - log (4)..Thanks tn4596, it makes more sense now.
I've attached a proof so that you can see how to actually solve for x. The rules for logarithmic operations aren't terribly complicated and it's probably not a bad idea to learn them. Hope this is helpful.
A previous responder tried to give you a rule for calculating logs, which can be somewhat misleading. I would simply learn how to use logarithms instead of trying to memorize a rule that isn't always applicable.
Also, not to be rude, but you can take the log of a negative number, it just winds up being complex and multi-valued. This is similar to how one can take the square root of a negative number and get a complex and multi-valued result (e.g., the square root of -4 equals +2i and -2i).
It's really very simple, and you don't need to take logs (or ln's).
2^x= 1/4 means "2 raised to what power equals 1/4. Obviously (1/2)^2=1/4. Reciprocal powers (1/n)^x are defined as n^-x, so that gives you the general answer. The answer in this particular case is "-2".
I agree. You don't - and shouldn't - need to use logs for this problem. But, I wanted to be complete since I had mentioned it in my original response.
I prefer logs, since i understand them better