OpenStudy (anonymous):

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

5 years ago
myininaya (myininaya):

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

5 years ago
myininaya (myininaya):

and z?

5 years ago
OpenStudy (anonymous):

In terms of y

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

2/y

5 years ago
OpenStudy (anonymous):

Can you explain?

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
myininaya (myininaya):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
myininaya (myininaya):

how did they do it?

5 years ago
myininaya (myininaya):

or they didn't?

5 years ago
OpenStudy (anonymous):

No solution

5 years ago
myininaya (myininaya):

i want to see ffm's explanation :)

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (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 ;)

5 years ago
OpenStudy (anonymous):

Thanks ffm

5 years ago
OpenStudy (turingtest):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (turingtest):

am I missing something?

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (turingtest):

ah, then FFM has done it again :D

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (turingtest):

what's JEE ?

5 years ago
OpenStudy (anonymous):

@FFM are you an IIT aspirant?

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

Except the last J2EE.

5 years ago
OpenStudy (anonymous):

This question was in a eigth grade book

5 years ago
OpenStudy (anonymous):

Lol, great :)

5 years ago
OpenStudy (anonymous):

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

5 years ago
myininaya (myininaya):

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

5 years ago
myininaya (myininaya):

nice job ffm :)

5 years ago
OpenStudy (anonymous):

Thanks myin :)

5 years ago
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