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

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so you want that in terms of a and b and c?
and z?
In terms of y

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Other answers:

(a) 2/y (b) 1/y (c) 1/2y (d) 2y
Can you explain?
Sure I can, tell me have you tried taking logarithms?
No, I don't think this question should be solved by logarithms
How do you think we should solve it? Are do you have any thoughts? Or is this a test for us?
I had this doubt from a book, under the topic algebra.
how did they do it?
or they didn't?
No solution
i want to see ffm's explanation :)
Is the question 1/x+1/z ? then it is 2/y
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 ;)
Thanks ffm
we wanted 1/x+1/y though you have 1/x+1/z....?
Glad to help :) and my apologies for the buggy Latex :(
am I missing something?
Sorry, it's (1/x)+(1/z)
ah, then FFM has done it again :D
lol, it's was pretty obvious, also it's a typical JEE problem :D
what's JEE ?
@FFM are you an IIT aspirant?
so now you are an iitian. I am writing the test this year
Turing, all of this:
Except the last J2EE.
This question was in a eigth grade book
Lol, great :)
That implies I am smarter than a 8 grader ? :P
lol i thought it weird to have a y in the expression to start with
nice job ffm :)
Thanks myin :)

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