Advertisements
Advertisements
प्रश्न
Find x, if : `(root(3)( 2/3))^( x - 1 ) = 27/8`
Advertisements
उत्तर
`(root(3)( 2/3))^( x - 1 ) = 27/8`
`[(2/3)^(1/3)]^( x - 1 ) = 3^3/2^3`
⇒ `(2/3)^[( x - 1 )/3] = (3/2)^3`
⇒ `(2/3)^[( x - 1 )/3] = (2/3)^-3`
We know that if bases are equal, the powers are equal
⇒ `[ x - 1 ]/3 = -3`
⇒ x - 1 = - 9
⇒ x = - 9 + 1
⇒ x = - 8
APPEARS IN
संबंधित प्रश्न
Solve : 4x - 2 - 2x + 1 = 0
Solve for x : 3(2x + 1) - 2x + 2 + 5 = 0
If 5-P = 4-q = 20r, show that : `1/p + 1/q + 1/r = 0`
Solve for x:
`sqrt((3/5)^(x + 3)) = (27^-1)/(125^-1)`
Solve for x:
9x+4 = 32 x (27)x+1
Find the value of k in each of the following:
`(1/3)^-4 ÷ 9^((-1)/(3)` = 3k
If `x^(1/3) + y^(1/3) + z^(1/3) = 0`, prove that (x + y + z)3 = 27xyz
If 2250 = 2a. 3b. 5c, find a, b and c. Hence, calculate the value of 3a x 2-b x 5-c.
If 2x = 3y = 12z ; show that `(1)/z = (1)/y + (2)/x`.
Find the value of 'a' and 'b' if:
`(sqrt243)^"a" ÷ 3^("b" + 1)` = 1 and `27^"b" - 81^(4 -"a"/2)` = 0
