Advertisements
Advertisements
Question
Find the value of x in the following:
`(root3 4)^(2x+1/2)=1/32`
Advertisements
Solution
Given `(root3 4)^(2x+1/2)=1/32`
`(2^2)^((1/3)((4x+1)/2))=(1/2)^5`
`rArr2^((4x+1)/3)=2^-5`
On comparing we get,
`(4x+1)/3=-5`
⇒ 4x + 1 = -5 x 3
⇒ 4x + 1 = -15
⇒ 4x = -15 - 1
⇒ 4x = -16
`rArrx=-16/4`
⇒ x = -4
Hence, the value of x = -4.
APPEARS IN
RELATED QUESTIONS
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`
Prove that:
`(a+b+c)/(a^-1b^-1+b^-1c^-1+c^-1a^-1)=abc`
Simplify:
`root3((343)^-2)`
If 3x = 5y = (75)z, show that `z=(xy)/(2x+y)`
Determine `(8x)^x,`If `9^(x+2)=240+9^x`
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`
The seventh root of x divided by the eighth root of x is
Which one of the following is not equal to \[\left( \frac{100}{9} \right)^{- 3/2}\]?
If \[\frac{x}{x^{1 . 5}} = 8 x^{- 1}\] and x > 0, then x =
If \[\frac{2^{m + n}}{2^{n - m}} = 16\], \[\frac{3^p}{3^n} = 81\] and \[a = 2^{1/10}\],than \[\frac{a^{2m + n - p}}{( a^{m - 2n + 2p} )^{- 1}} =\]
