Advertisements
Advertisements
प्रश्न
Solve for x:
22x+3 - 9 x 2x + 1 = 0
उत्तर
22x+3 - 9 x 2x + 1 = 0
22x . 23 - 9 x 2x + 1 = 0
Put2x = t, so, 22x = t2
So, 22x . 23 - 9 x 2x + 1 = 0 becomes 8t2 - 9t + 1 = 0
⇒ 8t2 - 8t - t + 1 = 0
⇒ 8t(t - 1) - (t - 1) = 0
⇒ t - 1 = 0 or 8t - 1 = 0
⇒ t = 1 or t = `(1)/(8)`
⇒ 2x = 1 or 2x = `(1)/(2^3)`
⇒ 2x = 20 or 2x = 2-3
⇒ x = 0 or x = -3.
APPEARS IN
संबंधित प्रश्न
Solve for x:
`2^(3x + 3) = 2^(3x + 1) + 48`
Evaluate : `4/(216)^(-2/3) + 1/(256)^(-3/4) + 2/(243)^(-1/5)`
Evaluate the following:
`(8/27)^((-2)/3) - (1/3)^-2 - 7^0`
Evaluate the following:
`16^(3/4) + 2(1/2)^-1 xx 3^0`
Solve for x:
1 = px
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:
`(sqrt(9))^-7 xx (sqrt(3))^-5` = 3k
If `root(x)("a") = root(y)("b") = root(z)("c")` and abc = 1, prove that x + y + z = 0
If 2x = 3y = 12z ; show that `(1)/z = (1)/y + (2)/x`.
Find the value of 'a' and 'b' if:
92a = `(root(3)(81))^(-6/"b") = (sqrt(27))^2`