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myininayaBest ResponseYou've already chosen the best response.2
so you want that in terms of a and b and c?
 2 years ago

abdul_shabeerBest ResponseYou've already chosen the best response.0
(a) 2/y (b) 1/y (c) 1/2y (d) 2y
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.5
Sure I can, tell me have you tried taking logarithms?
 2 years ago

abdul_shabeerBest ResponseYou've already chosen the best response.0
No, I don't think this question should be solved by logarithms
 2 years ago

myininayaBest ResponseYou've already chosen the best response.2
How do you think we should solve it? Are do you have any thoughts? Or is this a test for us?
 2 years ago

abdul_shabeerBest ResponseYou've already chosen the best response.0
I had this doubt from a book, under the topic algebra.
 2 years ago

myininayaBest ResponseYou've already chosen the best response.2
i want to see ffm's explanation :)
 2 years ago

vicky007Best ResponseYou've already chosen the best response.0
Is the question 1/x+1/z ? then it is 2/y
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.5
Okay without logarithms, Set \( a^x = b^y = c^z = k \) Then, \( a = k^{1/x}, b = k^{1/y}, c = k^{1/z} \) Given \( b^2 = ac \). So, \( (k^{1/y} )^2 = \( k^{1/x} \times k^{1/z} \) \( \Rightarrow k^{2/y} = k^{ (1/x) + (1/z)} \) Hence, 2/y = (1/x) + (1/z) QED ;)
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.0
we wanted 1/x+1/y though you have 1/x+1/z....?
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.5
Glad to help :) and my apologies for the buggy Latex :(
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.0
am I missing something?
 2 years ago

abdul_shabeerBest ResponseYou've already chosen the best response.0
Sorry, it's (1/x)+(1/z)
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.0
ah, then FFM has done it again :D
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.5
lol, it's was pretty obvious, also it's a typical JEE problem :D
 2 years ago

vicky007Best ResponseYou've already chosen the best response.0
@FFM are you an IIT aspirant?
 2 years ago

vicky007Best ResponseYou've already chosen the best response.0
so now you are an iitian. I am writing the test this year
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.5
Turing, all of this: http://en.wikipedia.org/wiki/JEE
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.5
Except the last J2EE.
 2 years ago

abdul_shabeerBest ResponseYou've already chosen the best response.0
This question was in a eigth grade book
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.5
That implies I am smarter than a 8 grader ? :P
 2 years ago

myininayaBest ResponseYou've already chosen the best response.2
lol i thought it weird to have a y in the expression to start with
 2 years ago
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