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
Simplify the following:
`(2x^-2y^3)^3`
Prove that:
`1/(1 + x^(b - a) + x^(c - a)) + 1/(1 + x^(a - b) + x^(c - b)) + 1/(1 + x^(b - c) + x^(a - c)) = 1`
Solve the following equations for x:
`2^(2x)-2^(x+3)+2^4=0`
Prove that:
`sqrt(3xx5^-3)divroot3(3^-1)sqrt5xxroot6(3xx5^6)=3/5`
Prove that:
`(64/125)^(-2/3)+1/(256/625)^(1/4)+(sqrt25/root3 64)=65/16`
If 2x = 3y = 6-z, show that `1/x+1/y+1/z=0`
Find the value of x in the following:
`(sqrt(3/5))^(x+1)=125/27`
When simplified \[\left( - \frac{1}{27} \right)^{- 2/3}\] is
Find:-
`32^(1/5)`
Find:-
`32^(2/5)`
