Advertisements
Advertisements
प्रश्न
Find the values of m and n if :
`4^(2m) = ( root(3)(16))^(-6/n) = (sqrt8)^2`
Advertisements
उत्तर
`4^(2"m") = ( root(3)(16))^(-6/"n") = (sqrt8)^2`
⇒ `4^(2"m") = (sqrt8)^2` ....(1)
and
`(root(3)(16))^(-6/n) = (sqrt8)^2` ....(2)
From (1)
`4^(2"m") = (sqrt8)^2`
⇒ `(2^2)^(2"m") = (sqrt(2^3))^2`
⇒ `2^(4"m") = [(2^3)^(1/2)]^2`
⇒ `2^(4"m") = [ 2^( 3 xx 1/2 )]^2`
⇒ `2^(4"m") = 2^( 3 xx 1/2 xx 2)`
⇒ `2^(4"m") = 2^3`
⇒ 4m = 3
⇒ m = `3/4`
From (2), We have
`(3sqrt(16))^(-6/"n") = (sqrt8)^2`
⇒ `( root(3)(2 xx 2 xx 2 xx 2))^(-6/"n") = (sqrt( 2 xx 2 xx 2))^2`
⇒ `( root(3)(2^4))^(-6/"n") = ( sqrt(2^3))^2`
⇒ `[(2^4)^(1/3)]^(-6/"n") = [(2^3)^(1/2)]^2`
⇒ `[2^(4/3)]^(-6/"n") = [2^(3/2)]^2`
⇒ `2^( 4/3 xx ( - 6/"n" ) = 2^(3/2 xx 2)`
⇒ `2^(-8/"n") = 2^3`
⇒ `-8/"n" = 3`
⇒ ` "n" = -8/3 "Thus m" = 3/4"n" = - 8/3`
APPEARS IN
संबंधित प्रश्न
Solve : `[3^x]^2` : 3x = 9 : 1
Solve:
22x + 2x+2 − 4 × 23 = 0
Solve for x : (a3x + 5)2. (ax)4 = a8x + 12
Solve for x:
`(81)^(3/4) - (1/32)^(-2/5) + x(1/2)^(-1).2^0 = 27`
Prove that :
`[ x^(a(b - c))]/[x^b(a - c)] ÷ ((x^b)/(x^a))^c = 1`
Evaluate the following:
`16^(3/4) + 2(1/2)^-1 xx 3^0`
Solve for x:
`9 xx 3^x = (27)^(2x - 5)`
Solve for x:
1 = px
If `root(x)("a") = root(y)("b") = root(z)("c")` and abc = 1, prove that x + y + z = 0
If 2400 = 2x x 3y x 5z, find the numerical value of x, y, z. Find the value of 2-x x 3y x 5z as fraction.
