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abdul_shabeer
 4 years ago
If a^x=b^y=c^z and b^2=ac, then (1/x)+(1/y) is
abdul_shabeer
 4 years ago
If a^x=b^y=c^z and b^2=ac, then (1/x)+(1/y) is

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

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

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

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

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

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

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

vicky007
 4 years ago
Best ResponseYou've already chosen the best response.0Is the question 1/x+1/z ? then it is 2/y

FoolForMath
 4 years ago
Best ResponseYou've already chosen the best response.5Okay 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 ;)

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

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0am I missing something?

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

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

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

vicky007
 4 years ago
Best ResponseYou've already chosen the best response.0@FFM are you an IIT aspirant?

vicky007
 4 years ago
Best ResponseYou've already chosen the best response.0so now you are an iitian. I am writing the test this year

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

FoolForMath
 4 years ago
Best ResponseYou've already chosen the best response.5Except the last J2EE.

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

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

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