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`
Prove that:
`(a+b+c)/(a^-1b^-1+b^-1c^-1+c^-1a^-1)=abc`
Solve the following equation for x:
`2^(x+1)=4^(x-3)`
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrtx)^((-2)/3)sqrt(y^4)divsqrt(xy^((-1)/2))`
Simplify:
`(16^(-1/5))^(5/2)`
If a and b are different positive primes such that
`((a^-1b^2)/(a^2b^-4))^7div((a^3b^-5)/(a^-2b^3))=a^xb^y,` find x and y.
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 − 1)3 = 8, What is the value of (x + 1)2 ?
Find:-
`125^(1/3)`
Find:-
`125^((-1)/3)`
