Advertisements
Advertisements
Question
If \[\sqrt{2^n} = 1024,\] then \[{3^2}^\left( \frac{n}{4} - 4 \right) =\]
Options
3
9
27
81
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 abc = 1, show that `1/(1+a+b^-1)+1/(1+b+c^-1)+1/(1+c+a^-1)=1`
If 1176 = `2^axx3^bxx7^c,` find the values of a, b and c. Hence, compute the value of `2^axx3^bxx7^-c` as a fraction.
If `x = a^(m+n),` `y=a^(n+l)` and `z=a^(l+m),` prove that `x^my^nz^l=x^ny^lz^m`
Write \[\left( 625 \right)^{- 1/4}\] in decimal form.
Write \[\left( \frac{1}{9} \right)^{- 1/2} \times (64 )^{- 1/3}\] as a rational number.
Write the value of \[\left\{ 5( 8^{1/3} + {27}^{1/3} )^3 \right\}^{1/4} . \]
The product of the square root of x with the cube root of x is
If (23)2 = 4x, then 3x =
If x = \[\sqrt[3]{2 + \sqrt{3}}\] , then \[x^3 + \frac{1}{x^3} =\]
Simplify:
`11^(1/2)/11^(1/4)`
