Question

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

Answer 1

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

Answer 2

and z?

Answer 3

In terms of y

Answer 4

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

Answer 5

2/y

Answer 6

Can you explain?

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

how did they do it?

Answer 12

or they didn't?

Answer 13

No solution

Answer 14

i want to see ffm's explanation :)

Answer 15

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

Answer 16

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 ;)

Answer 17

Thanks ffm

Answer 18

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

Answer 19

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

Answer 20

am I missing something?

Answer 21

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

Answer 22

ah, then FFM has done it again :D

Answer 23

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

Answer 24

what's JEE ?

Answer 25

@FFM are you an IIT aspirant?

Answer 26

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

Answer 27

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

Answer 28

Except the last J2EE.

Answer 29

This question was in a eigth grade book

Answer 30

Lol, great :)

Answer 31

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

Answer 32

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

Answer 33

nice job ffm :)

Answer 34

Thanks myin :)

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