Advertisements
Advertisements
Question
If `root(x)("a") = root(y)("b") = root(z)("c")` and abc = 1, prove that x + y + z = 0
Advertisements
Solution
Let `root(x)("a") = root(y)("b") = root(z)("c")`
⇒ `"a"^(1/x) = "k", "b"^(1/y) = "k", "c"^(1/z) = "k"`
⇒ a = k, b = k, c = k
It is also given that abc = 1
⇒ kx x ky x kz = 1
⇒ `"k"^(x + y + z)` = k°
⇒ x + y + z = 0.
APPEARS IN
RELATED QUESTIONS
Find x, if : `(root(3)( 2/3))^( x - 1 ) = 27/8`
Solve for x:
`2^(3x + 3) = 2^(3x + 1) + 48`
If 5-P = 4-q = 20r, show that : `1/p + 1/q + 1/r = 0`
Evaluate the following:
`(2^3 xx 3^5 xx 24^2)/(12^2 xx 18^3 xx 27)`
Solve for x:
3 x 7x = 7 x 3x
Solve for x:
9 x 81x = `(1)/(27^(x - 3)`
Solve for x:
22x- 1 − 9 x 2x − 2 + 1 = 0
Find the value of k in each of the following:
`(root(3)(8))^((-1)/(2)` = 2k
Find the value of k in each of the following:
`(1/3)^-4 ÷ 9^((-1)/(3)` = 3k
If a = `2^(1/3) - 2^((-1)/3)`, prove that 2a3 + 6a = 3
