Advertisements
Advertisements
प्रश्न
If ax = by = cz and b2 = ac, prove that y = `(2xz)/(z + x)`
Advertisements
उत्तर
Let ax = by = cz = k
⇒ `"a" = "k"^(1/x), "b" = "k"^(1/y), "c" = "k"^(1/2)`
It is also given that b2 = ac
⇒ `"k"^(2/y) = "k"^(1/x) xx "k"^(1/2)`
⇒ `"k"^(2/y) = "k"^(1/x + 1/z)`
⇒ `(2)/y = (1)/x + (1)/z`
⇒ y = `(2zx)/(z + x)`.
APPEARS IN
संबंधित प्रश्न
Solve for x:
`3^(4x + 1) = (27)^(x + 1)`
Solve for x : (a3x + 5)2. (ax)4 = a8x + 12
Solve for x : 3(2x + 1) - 2x + 2 + 5 = 0
Prove that :
`[ x^(a(b - c))]/[x^b(a - c)] ÷ ((x^b)/(x^a))^c = 1`
If `((a^-1b^2 )/(a^2b^-4))^7 ÷ (( a^3b^-5)/(a^-2b^3))^-5 = a^x . b^y` , find x + y.
Evaluate the following:
`sqrt(1/4) + (0.01)^(-1/2) - (27)^(2/3)`
Solve for x:
22x+1= 8
Find the value of k in each of the following:
`root(4)root(3)(x^2)` = xk
If a = `2^(1/3) - 2^((-1)/3)`, prove that 2a3 + 6a = 3
If `x^(1/3) + y^(1/3) + z^(1/3) = 0`, prove that (x + y + z)3 = 27xyz
