## abdul_shabeer 3 years ago If a^x=b^y=c^z and b^2=ac, then (1/x)+(1/y) is

1. myininaya

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

2. myininaya

and z?

3. abdul_shabeer

In terms of y

4. abdul_shabeer

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

5. FoolForMath

2/y

6. abdul_shabeer

Can you explain?

7. FoolForMath

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

8. abdul_shabeer

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

9. myininaya

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

10. abdul_shabeer

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

11. myininaya

how did they do it?

12. myininaya

or they didn't?

13. abdul_shabeer

No solution

14. myininaya

i want to see ffm's explanation :)

15. vicky007

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

16. FoolForMath

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 ;)

17. abdul_shabeer

Thanks ffm

18. TuringTest

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

19. FoolForMath

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

20. TuringTest

am I missing something?

21. abdul_shabeer

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

22. TuringTest

ah, then FFM has done it again :D

23. FoolForMath

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

24. TuringTest

what's JEE ?

25. vicky007

@FFM are you an IIT aspirant?

26. vicky007

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

27. FoolForMath

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

28. FoolForMath

Except the last J2EE.

29. abdul_shabeer

This question was in a eigth grade book

30. FoolForMath

Lol, great :)

31. FoolForMath

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

32. myininaya

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

33. myininaya

nice job ffm :)

34. FoolForMath

Thanks myin :)