Advertisements
Advertisements
प्रश्न
Find:-
`32^(2/5)`
Advertisements
उत्तर
We can write the given expression as follows
⇒ `32^(2/5) = (2^5)^(2/5)`
On simplifying
⇒ `32^(2/5) = 2^(5 xx 2/5)`
⇒ `32^(2/5) = 2^2`
∴ `32^(2/5) = 4`
APPEARS IN
संबंधित प्रश्न
If `1176=2^a3^b7^c,` find a, b and c.
Simplify:
`(sqrt2/5)^8div(sqrt2/5)^13`
Show that:
`[{x^(a(a-b))/x^(a(a+b))}div{x^(b(b-a))/x^(b(b+a))}]^(a+b)=1`
Find the value of x in the following:
`(2^3)^4=(2^2)^x`
If `3^(4x) = (81)^-1` and `10^(1/y)=0.0001,` find the value of ` 2^(-x+4y)`.
Simplify:
`(x^(a+b)/x^c)^(a-b)(x^(b+c)/x^a)^(b-c)(x^(c+a)/x^b)^(c-a)`
For any positive real number x, write the value of \[\left\{ \left( x^a \right)^b \right\}^\frac{1}{ab} \left\{ \left( x^b \right)^c \right\}^\frac{1}{bc} \left\{ \left( x^c \right)^a \right\}^\frac{1}{ca}\]
When simplified \[(256) {}^{- ( 4^{- 3/2} )}\] is
If \[\sqrt{5^n} = 125\] then `5nsqrt64`=
If \[\sqrt{13 - a\sqrt{10}} = \sqrt{8} + \sqrt{5}, \text { then a } =\]
