Advertisements
Advertisements
प्रश्न
Write the value of \[\sqrt[3]{7} \times \sqrt[3]{49} .\]
Advertisements
उत्तर
We have to find the value of . `3sqrt7 xx 3sqrt49.`So,
`3sqrt7 xx 3sqrt49 = 3sqrt7 xx 3sqrt49.`
`= 7^(1/3) xx 7^(2 xx 1/3)`
`= 7^(1/3) xx 7^(2 /3)`
By using law rational exponents `a^m xx a^n = a^(m+n)` we get,
`3sqrt7 xx 3sqrt49 =7^(1/3) xx 7^(2 /3)`
`= 7^(1/3+2/3 )`
`= 7^(3/3)` = 7
Hence the value of `3sqrt7 xx 3sqrt49` is 7
APPEARS IN
संबंधित प्रश्न
If `a=xy^(p-1), b=xy^(q-1)` and `c=xy^(r-1),` prove that `a^(q-r)b^(r-p)c^(p-q)=1`
Simplify:
`((25)^(3/2)xx(243)^(3/5))/((16)^(5/4)xx(8)^(4/3))`
Show that:
`(3^a/3^b)^(a+b)(3^b/3^c)^(b+c)(3^c/3^a)^(c+a)=1`
If 2x = 3y = 6-z, show that `1/x+1/y+1/z=0`
If `27^x=9/3^x,` find x.
If a, b, c are positive real numbers, then \[\sqrt[5]{3125 a^{10} b^5 c^{10}}\] is equal to
The value of m for which \[\left[ \left\{ \left( \frac{1}{7^2} \right)^{- 2} \right\}^{- 1/3} \right]^{1/4} = 7^m ,\] is
\[\frac{1}{\sqrt{9} - \sqrt{8}}\] is equal to
The positive square root of \[7 + \sqrt{48}\] is
Find:-
`125^((-1)/3)`
