Advertisements
Advertisements
Question
If \[\sqrt{2^n} = 1024,\] then \[{3^2}^\left( \frac{n}{4} - 4 \right) =\]
Options
3
9
27
81
Advertisements
Solution
We have to find `3^(2(n/4-4))`
Given `sqrt(2^n) = 1024`
`(sqrt(2^n) = 2^10`
`2^(nxx1/2) = 2^10`
Equating powers of rational exponents we get
`n xx 1/2 = 10`
`n = 10 xx 2`
`n =20`
Substituting in `3^(2(n/4-4))` ``we get
`3^(2(n/4-4)) = 3^(2(20/4-4))`
`= 3^(2(5-4))`
` =3^(2xx1)`
`= 9`
APPEARS IN
RELATED QUESTIONS
Simplify the following:
`(2x^-2y^3)^3`
Simplify the following
`((x^2y^2)/(a^2b^3))^n`
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrtx)^((-2)/3)sqrt(y^4)divsqrt(xy^((-1)/2))`
Simplify:
`((25)^(3/2)xx(243)^(3/5))/((16)^(5/4)xx(8)^(4/3))`
Determine `(8x)^x,`If `9^(x+2)=240+9^x`
Simplify:
`(x^(a+b)/x^c)^(a-b)(x^(b+c)/x^a)^(b-c)(x^(c+a)/x^b)^(c-a)`
If \[\frac{x}{x^{1 . 5}} = 8 x^{- 1}\] and x > 0, then x =
If \[\sqrt{5^n} = 125\] then `5nsqrt64`=
If \[x = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}\] and \[y = \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}\] then x + y +xy=
If x= \[\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\] and y = \[\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}\] , then x2 + y +y2 =
