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
Solve the following equation for x:
`2^(3x-7)=256`
Assuming that x, y, z are positive real numbers, simplify the following:
`sqrt(x^3y^-2)`
Solve the following equation:
`8^(x+1)=16^(y+2)` and, `(1/2)^(3+x)=(1/4)^(3y)`
The value of \[\left\{ 2 - 3 (2 - 3 )^3 \right\}^3\] is
Which of the following is (are) not equal to \[\left\{ \left( \frac{5}{6} \right)^{1/5} \right\}^{- 1/6}\] ?
The value of \[\left\{ 8^{- 4/3} \div 2^{- 2} \right\}^{1/2}\] is
If g = `t^(2/3) + 4t^(-1/2)`, what is the value of g when t = 64?
If \[\frac{2^{m + n}}{2^{n - m}} = 16\], \[\frac{3^p}{3^n} = 81\] and \[a = 2^{1/10}\],than \[\frac{a^{2m + n - p}}{( a^{m - 2n + 2p} )^{- 1}} =\]
If \[\sqrt{2^n} = 1024,\] then \[{3^2}^\left( \frac{n}{4} - 4 \right) =\]
Simplify:-
`(1/3^3)^7`
