Question

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

Answer 1

1 sec

Answer 2

first you have to choose a variable to eliminate

Answer 3

can u gimme the proper solution for this

Answer 4

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

Answer 5

hang on looking in my book

Answer 6

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

, solve it accordingly

Answer 7

algebra book?

Answer 8

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

Answer 9

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.

Answer 10

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

Answer 11

okay, i'll try to find something ^^

Answer 12

thanks to all who r trying

Answer 13

still working on it

Answer 14

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.

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

try drawing it to make sure I understand your notation

Answer 19

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

Answer 20

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

Answer 21

what class is this from csr21?

Answer 22

this is from class 9 cbse board

Answer 23

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

Answer 24

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

Answer 25

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.

Answer 26

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 :
\[({\ln 2 + \ln 3 - \ln 6}) /{ \ln 2 * x} = 0\]

Answer 27

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