OpenStudy (anonymous):

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

6 years ago
myininaya (myininaya):

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

6 years ago
myininaya (myininaya):

and z?

6 years ago
OpenStudy (anonymous):

In terms of y

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

2/y

6 years ago
OpenStudy (anonymous):

Can you explain?

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

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

6 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?

6 years ago
OpenStudy (anonymous):

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

6 years ago
myininaya (myininaya):

how did they do it?

6 years ago
myininaya (myininaya):

or they didn't?

6 years ago
OpenStudy (anonymous):

No solution

6 years ago
myininaya (myininaya):

i want to see ffm's explanation :)

6 years ago
OpenStudy (anonymous):

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

6 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 ;)

6 years ago
OpenStudy (anonymous):

Thanks ffm

6 years ago
OpenStudy (turingtest):

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

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (turingtest):

am I missing something?

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (turingtest):

ah, then FFM has done it again :D

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (turingtest):

what's JEE ?

6 years ago
OpenStudy (anonymous):

@FFM are you an IIT aspirant?

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

Except the last J2EE.

6 years ago
OpenStudy (anonymous):

This question was in a eigth grade book

6 years ago
OpenStudy (anonymous):

Lol, great :)

6 years ago
OpenStudy (anonymous):

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

6 years ago
myininaya (myininaya):

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

6 years ago
myininaya (myininaya):

nice job ffm :)

6 years ago
OpenStudy (anonymous):

Thanks myin :)

6 years ago