Advertisements
Advertisements
प्रश्न
If ax = by = cz and b2 = ac, show that `y=(2zx)/(z+x)`
उत्तर
Let ax = by = cz = k
So, `a=k^(1/x),` `b=k^(1/y),` c=k^(1/z)
Thus,
`b^2 = ac`
`rArr(k^(1/y))^2=(k^(1/x))(k^(1/z))`
`rArrk^(2/y)=k^(1/x+1/z)`
`rArr2/y=1/x+1/z`
`rArr2/y=(z+x)/(xz)`
`rArr2xx(zx)/(z+x)=y`
`rArry=(2zx)/(z+x)`
APPEARS IN
संबंधित प्रश्न
If a = 3 and b = -2, find the values of :
ab + ba
Simplify:
`((5^-1xx7^2)/(5^2xx7^-4))^(7/2)xx((5^-2xx7^3)/(5^3xx7^-5))^(-5/2)`
Prove that:
`9^(3/2)-3xx5^0-(1/81)^(-1/2)=15`
Find the value of x in the following:
`(13)^(sqrtx)=4^4-3^4-6`
Find the value of x in the following:
`(sqrt(3/5))^(x+1)=125/27`
Solve the following equation:
`3^(x-1)xx5^(2y-3)=225`
Write the value of \[\left\{ 5( 8^{1/3} + {27}^{1/3} )^3 \right\}^{1/4} . \]
If (x − 1)3 = 8, What is the value of (x + 1)2 ?
If \[\sqrt{5^n} = 125\] then `5nsqrt64`=
The value of \[\sqrt{3 - 2\sqrt{2}}\] is
