Advertisements
Advertisements
प्रश्न
Solve : `[3^x]^2` : 3x = 9 : 1
उत्तर
`[3^x]^2` : 3x = 9 : 1
⇒ `[3^x]^2/3^x = 9/1`
⇒ `[3^x]^2 = 9 xx 3^x`
⇒ `[3^x]^2 = 3^2 xx 3^x`
⇒ `[3^x]^2 = 3^(x + 2)`
We know that if bases are equal, the powers are equal.
⇒ x2 = x + 2
⇒ x2 - x - 2 = 0
⇒ x2 - (2 - 1) x - 2 = 0
⇒ x2 - 2x + x - 2 = 0
⇒ x( x - 2 ) + 1( x - 2 ) = 0
⇒ ( x + 1 )( x - 2 ) = 0
⇒ x + 1 = 0 or x - 2 = 0
⇒ x = - 1 or x = 2.
APPEARS IN
संबंधित प्रश्न
Solve for x : 25x-1 = 4 23x + 1
Find x, if : 42x = `1/32`
Find x, if : `(root(3)( 2/3))^( x - 1 ) = 27/8`
Solve for x: `4^(x-1) × (0.5)^(3 - 2x) = (1/8)^-x`
Prove that :
`[ x^(a(b - c))]/[x^b(a - c)] ÷ ((x^b)/(x^a))^c = 1`
Evaluate the following:
`(12^2 xx 75^-2 xx 35 xx 400)/(48^2 xx 15^-3 xx 525)`
Evaluate the following:
`(27)^(2/3) xx 8^((-1)/6) ÷ 18^((-1)/2)`
Solve for x:
`sqrt((8^0 + 2/3)` = (0.6)2-3x
Find the value of k in each of the following:
`(root(3)(8))^((-1)/(2)` = 2k
Find the value of (8p)p if 9p + 2 - 9p = 240.