Advertisements
Advertisements
प्रश्न
If \[\sqrt{5^n} = 125\] then `5nsqrt64`=
पर्याय
25
\[\frac{1}{125}\]
625
\[\frac{1}{5}\]
Advertisements
उत्तर
We have to find `5nsqrt64` provided \[\sqrt{5^n} = 125\]
So,
`sqrt 5^n = 125`
`5^(nxx 1/2)= 5^3`
`n/2 = 3`
`n=3xx2`
` n =6`
Substitute ` n =6` in `5nsqrt64` to get
` `5nsqrt64 = 5^(2^(6x1/6)`
=` 5^(2^(6x1/6)`
`= 5xx5`
`=25`
Hence the value of `5nsqrt64` is 25.
APPEARS IN
संबंधित प्रश्न
Prove that:
`(x^a/x^b)^(a^2+ab+b^2)xx(x^b/x^c)^(b^2+bc+c^2)xx(x^c/x^a)^(c^2+ca+a^2)=1`
Assuming that x, y, z are positive real numbers, simplify the following:
`(x^-4/y^-10)^(5/4)`
Prove that:
`(2^n+2^(n-1))/(2^(n+1)-2^n)=3/2`
Show that:
`(x^(a^2+b^2)/x^(ab))^(a+b)(x^(b^2+c^2)/x^(bc))^(b+c)(x^(c^2+a^2)/x^(ac))^(a+c)=x^(2(a^3+b^3+c^3))`
If 2x = 3y = 6-z, show that `1/x+1/y+1/z=0`
If ax = by = cz and b2 = ac, show that `y=(2zx)/(z+x)`
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, m, n are positive ingegers, then \[\left\{ \sqrt[m]{\sqrt[n]{a}} \right\}^{mn}\] is equal to
If 9x+2 = 240 + 9x, then x =
If \[4x - 4 x^{- 1} = 24,\] then (2x)x equals
