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
If a = 3 and b = -2, find the values of :
ab + ba
Solve the following equations for x:
`2^(2x)-2^(x+3)+2^4=0`
Solve the following equations for x:
`3^(2x+4)+1=2.3^(x+2)`
Show that:
`[{x^(a(a-b))/x^(a(a+b))}div{x^(b(b-a))/x^(b(b+a))}]^(a+b)=1`
For any positive real number x, write the value of \[\left\{ \left( x^a \right)^b \right\}^\frac{1}{ab} \left\{ \left( x^b \right)^c \right\}^\frac{1}{bc} \left\{ \left( x^c \right)^a \right\}^\frac{1}{ca}\]
Which one of the following is not equal to \[\left( \sqrt[3]{8} \right)^{- 1/2} ?\]
If \[\frac{x}{x^{1 . 5}} = 8 x^{- 1}\] and x > 0, then x =
If (16)2x+3 =(64)x+3, then 42x-2 =
If o <y <x, which statement must be true?
The value of \[\sqrt{5 + 2\sqrt{6}}\] is
