Advertisements
Advertisements
Question
If x-2 = 64, then x1/3+x0 =
Options
2
3
3/2
2/3
Advertisements
Solution
We have to find the value of `x^(1/3+ x^0)` if `x^-2 = 64`
Consider,
`x^-2 = 2^6`
`1/x^2 = 2^6`
Multiply `1/2`on both sides of powers we get
`1/x^(2 xx 1/2) = 2^(6x1/2)`
`1/x^(2 xx 1/2) = 2^(6x1/2)`
`1/x = 2^3`
`1/x = 8/ 1`
By taking reciprocal on both sides we get,
`1/8 = x `
Substituting `1/8` in `x^(1/3 + x^0)`we get
`= (1/8)^(1/3) + (1/8)^0`
`= (1/(2^3))^(1/3) + (1/8)^0`
`= 1/(2^(3xx 1/3)) +1`
` = 1/2^1 + 1`
`=1/2 +1`
By taking least common multiply we get
`= 1/2 + (1 xx 2)/(1 xx 2)`
` = 1/2 + 2/2 `
`= (1+2)/2`
`= 3/2`
APPEARS IN
RELATED QUESTIONS
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrt(x^-3))^5`
Simplify:
`((25)^(3/2)xx(243)^(3/5))/((16)^(5/4)xx(8)^(4/3))`
Prove that:
`(3^-3xx6^2xxsqrt98)/(5^2xxroot3(1/25)xx(15)^(-4/3)xx3^(1/3))=28sqrt2`
Find the value of x in the following:
`5^(2x+3)=1`
If a and b are different positive primes such that
`(a+b)^-1(a^-1+b^-1)=a^xb^y,` find x + y + 2.
If `a=x^(m+n)y^l, b=x^(n+l)y^m` and `c=x^(l+m)y^n,` Prove that `a^(m-n)b^(n-l)c^(l-m)=1`
Simplify \[\left[ \left\{ \left( 625 \right)^{- 1/2} \right\}^{- 1/4} \right]^2\]
The value of \[\left\{ 2 - 3 (2 - 3 )^3 \right\}^3\] is
Which of the following is (are) not equal to \[\left\{ \left( \frac{5}{6} \right)^{1/5} \right\}^{- 1/6}\] ?
When simplified \[\left( - \frac{1}{27} \right)^{- 2/3}\] is
