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:
`(5xx25^(n+1)-25xx5^(2n))/(5xx5^(2n+3)-25^(n+1))`
Simplify:
`(16^(-1/5))^(5/2)`
Simplify:
`root3((343)^-2)`
If 3x = 5y = (75)z, show that `z=(xy)/(2x+y)`
Find the value of x in the following:
`2^(x-7)xx5^(x-4)=1250`
Find the value of x in the following:
`(13)^(sqrtx)=4^4-3^4-6`
`(2/3)^x (3/2)^(2x)=81/16 `then x =
The value of \[\frac{\sqrt{48} + \sqrt{32}}{\sqrt{27} + \sqrt{18}}\] is
If \[\sqrt{2} = 1 . 4142\] then \[\sqrt{\frac{\sqrt{2} - 1}{\sqrt{2} + 1}}\] is equal to
Find:-
`125^((-1)/3)`
