Mathematics
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OpenStudy (anonymous):
If a^x=b^y=c^z and b^2=ac, then (1/x)+(1/y) is
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myininaya (myininaya):
so you want that in terms of a and b and c?
myininaya (myininaya):
and z?
OpenStudy (anonymous):
In terms of y
OpenStudy (anonymous):
(a) 2/y (b) 1/y
(c) 1/2y (d) 2y
OpenStudy (anonymous):
2/y
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OpenStudy (anonymous):
Can you explain?
OpenStudy (anonymous):
Sure I can, tell me have you tried taking logarithms?
OpenStudy (anonymous):
No, I don't think this question should be solved by logarithms
myininaya (myininaya):
How do you think we should solve it? Are do you have any thoughts? Or is this a test for us?
OpenStudy (anonymous):
I had this doubt from a book, under the topic algebra.
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myininaya (myininaya):
how did they do it?
myininaya (myininaya):
or they didn't?
OpenStudy (anonymous):
No solution
myininaya (myininaya):
i want to see ffm's explanation :)
OpenStudy (anonymous):
Is the question 1/x+1/z ? then it is 2/y
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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 ;)
OpenStudy (anonymous):
Thanks ffm
OpenStudy (turingtest):
we wanted 1/x+1/y though
you have 1/x+1/z....?
OpenStudy (anonymous):
Glad to help :) and my apologies for the buggy Latex :(
OpenStudy (turingtest):
am I missing something?
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OpenStudy (anonymous):
Sorry, it's (1/x)+(1/z)
OpenStudy (turingtest):
ah, then FFM has done it again :D
OpenStudy (anonymous):
lol, it's was pretty obvious, also it's a typical JEE problem :D
OpenStudy (turingtest):
what's JEE ?
OpenStudy (anonymous):
@FFM are you an IIT aspirant?
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OpenStudy (anonymous):
so now you are an iitian. I am writing the test this year
OpenStudy (anonymous):
Except the last J2EE.
OpenStudy (anonymous):
This question was in a eigth grade book
OpenStudy (anonymous):
Lol, great :)
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OpenStudy (anonymous):
That implies I am smarter than a 8 grader ? :P
myininaya (myininaya):
lol i thought it weird to have a y in the expression to start with
myininaya (myininaya):
nice job ffm :)
OpenStudy (anonymous):
Thanks myin :)