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 for x:
`3^(4x + 1) = (27)^(x + 1)`
Find x, if : `( sqrt(3/5))^( x + 1) = 125/27`
Solve : `(sqrt(3))^( x - 3 ) = ( root(4)(3))^( x + 1 )`
Solve for x:
`2^(3x + 3) = 2^(3x + 1) + 48`
Simplify : `[ 3 xx 9^( n + 1 ) - 9 xx 3^(2n)]/[3 xx 3^(2n + 3) - 9^(n + 1 )]`
Evaluate the following:
`(2^6 xx 5^-4 xx 3^-3 xx 4^2)/(8^3 xx 15^-3 xx 25^-1)`
Solve for x:
`"p"^-5 = (1)/"p"^(x + 1)`
Solve for x:
5x2 : 5x = 25 : 1
If a = `2^(1/3) - 2^((-1)/3)`, prove that 2a3 + 6a = 3
Find the value of (8p)p if 9p + 2 - 9p = 240.
