if 2^x = 3^y = 6^-z, show that 1/x + 1/y + 1/z = 0

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first you have to choose a variable to eliminate

can u gimme the proper solution for this

not yet and it's driving me crazy, I'm a little rusty. I know this, just can't remember how to convert the exponent variables

hang on looking in my book

well i think i should inform u that its 9th standard problem in chapter "Number System" , solve it accordingly

algebra book?

"Number System " Chapter where rational and irrational numbers are there

one thing you can figure is : try to use the fact that 6 = 3*2 to have only 3 and 2 in your first equation.

thanks Reku, i did it , still i need the full solution now dear.. coz it dint really work

okay, i'll try to find something ^^

thanks to all who r trying

still working on it

I think I can solve it if I can convert the first part to fractions for example 6^-z=1/6^z but I can't remember how to do the 2^x and 3^y but I know it's simple.

actually i've also dont it , i dont know whether my method is correct or not. lemme post it here . u guys decide whether i m correct or not... 2^x=3^y=6^(-z) , this implies 2^x=3^y=(2.3)^(-z) 2^x=2^(-z).3^(-z) or i can write, 2^x.3^0=2^(-z).3^(-z) so i can write x=-z and z=0 similarly, i can solve for y and we get y=-z and z=0 Now, come to the equation 1/x + 1/y +1/z = 1/(-z) + 1/(-z) + 1/z =-1/z - 1/z + 1/z =-1/z = 0 (as z=0) hence proved please tell me whether the way i go is correct or not

hum 1/z doesn't exist if z=0

oops yesss u r correct reku, thats right .. so my solution is not correct

try drawing it to make sure I understand your notation

but if it is algebra, there is something in your solution that can maybe help us

what are x, y and z ? integer or float ?

what class is this from csr21?

this is from class 9 cbse board

nothing such is given, x,y and are integer or float or what?- nothing like that is given

um, why dont you try using logarithms. it would all become much simpler

As dhatraditya said, just write x ln 2 = y ln3 etc and sub in the rhs for y and z, say, and it will come out nicely.

lol it was so simple :\[2^x = 3^y <=> \ln 2 * x = \ln 3 * y \] \[6^{-z} = 2^x <=> \ln 6 * z = - \ln 2 * x\] so you have now z and y in function of x and the final expression becomes : \[1/x + \ln 3 / (\ln 2*x) + \ln 6 / (\ln2 * x)\] finaly : \[({\ln 2 + \ln 3 - \ln 6}) /{ \ln 2 * x} = 0\]

(because 2^x = exp(x *ln 2) ) (it's the definition)

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