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



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myininaya
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so you want that in terms of a and b and c?

myininaya
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and z?

abdul_shabeer
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In terms of y

abdul_shabeer
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(a) 2/y (b) 1/y
(c) 1/2y (d) 2y

FoolForMath
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2/y

abdul_shabeer
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Can you explain?

FoolForMath
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Sure I can, tell me have you tried taking logarithms?

abdul_shabeer
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No, I don't think this question should be solved by logarithms

myininaya
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How do you think we should solve it? Are do you have any thoughts? Or is this a test for us?

abdul_shabeer
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I had this doubt from a book, under the topic algebra.

myininaya
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how did they do it?

myininaya
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or they didn't?

abdul_shabeer
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No solution

myininaya
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i want to see ffm's explanation :)

vicky007
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Is the question 1/x+1/z ? then it is 2/y

FoolForMath
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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 ;)

abdul_shabeer
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Thanks ffm

TuringTest
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we wanted 1/x+1/y though
you have 1/x+1/z....?

FoolForMath
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Glad to help :) and my apologies for the buggy Latex :(

TuringTest
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am I missing something?

abdul_shabeer
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Sorry, it's (1/x)+(1/z)

TuringTest
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ah, then FFM has done it again :D

FoolForMath
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lol, it's was pretty obvious, also it's a typical JEE problem :D

TuringTest
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what's JEE ?

vicky007
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@FFM are you an IIT aspirant?

vicky007
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so now you are an iitian. I am writing the test this year


FoolForMath
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Except the last J2EE.

abdul_shabeer
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This question was in a eigth grade book

FoolForMath
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Lol, great :)

FoolForMath
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That implies I am smarter than a 8 grader ? :P

myininaya
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lol i thought it weird to have a y in the expression to start with

myininaya
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nice job ffm :)

FoolForMath
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Thanks myin :)