Advertisements
Advertisements
प्रश्न
If 3x = 5y = (75)z, show that `z=(xy)/(2x+y)`
Advertisements
उत्तर
Let 3x = 5y = (75)z = k
`rArr3=k^(1/x),` `5=k^(1/y),` `75=k^(1/z)`
`rArr5^2xx3=k^(1/z)`
`rArr(k^(1/y))^2xxk^(1/x)=k^(1/z)`
`rArrk^(2/y)xxk^(1/x)=k^(1/z)`
`rArrk^(2/y+1/x)=k^(1/z)`
`rArr2/y+1/x=1/z`
`rArr(2x+y)/(xy)=1/z`
`rArrz=(xy)/(2x+y)`
APPEARS IN
संबंधित प्रश्न
Simplify the following
`((4xx10^7)(6xx10^-5))/(8xx10^4)`
Simplify the following:
`(3^nxx9^(n+1))/(3^(n-1)xx9^(n-1))`
Simplify:
`root3((343)^-2)`
Prove that:
`((0.6)^0-(0.1)^-1)/((3/8)^-1(3/2)^3+((-1)/3)^-1)=(-3)/2`
If 2x = 3y = 12z, show that `1/z=1/y+2/x`
Find the value of x in the following:
`(13)^(sqrtx)=4^4-3^4-6`
If `3^(4x) = (81)^-1` and `10^(1/y)=0.0001,` find the value of ` 2^(-x+4y)`.
If 3x-1 = 9 and 4y+2 = 64, what is the value of \[\frac{x}{y}\] ?
(256)0.16 × (256)0.09
Find:-
`125^((-1)/3)`
