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

- anonymous

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

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- myininaya

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

- myininaya

and z?

- anonymous

In terms of y

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## More answers

- anonymous

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

- anonymous

2/y

- anonymous

Can you explain?

- anonymous

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

- anonymous

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

- myininaya

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

- anonymous

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

- myininaya

how did they do it?

- myininaya

or they didn't?

- anonymous

No solution

- myininaya

i want to see ffm's explanation :)

- anonymous

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

- anonymous

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 ;)

- anonymous

Thanks ffm

- TuringTest

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

- anonymous

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

- TuringTest

am I missing something?

- anonymous

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

- TuringTest

ah, then FFM has done it again :D

- anonymous

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

- TuringTest

what's JEE ?

- anonymous

@FFM are you an IIT aspirant?

- anonymous

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

- anonymous

Turing, all of this: http://en.wikipedia.org/wiki/JEE

- anonymous

Except the last J2EE.

- anonymous

This question was in a eigth grade book

- anonymous

Lol, great :)

- anonymous

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

- myininaya

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

- myininaya

nice job ffm :)

- anonymous

Thanks myin :)

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