Advertisements
Advertisements
प्रश्न
Find the value of x in the following:
`(sqrt(3/5))^(x+1)=125/27`
Advertisements
उत्तर
Given `(sqrt(3/5))^(x+1)=125/27`
`(sqrt(3/5))^(x+1)=(5/3)^3`
`rArr(3/5)^((x+1)/2)=(3/5)^-3`
On comparing we get,
`(x+1)/2=-3`
⇒ x + 1 = -3 x 2
⇒ x + 1 = -6
⇒ x = -6 - 1
⇒ x = -7
Hence, the value of x = -7.
APPEARS IN
संबंधित प्रश्न
Assuming that x, y, z are positive real numbers, simplify the following:
`(x^-4/y^-10)^(5/4)`
Simplify:
`root5((32)^-3)`
The square root of 64 divided by the cube root of 64 is
If x = 2 and y = 4, then \[\left( \frac{x}{y} \right)^{x - y} + \left( \frac{y}{x} \right)^{y - x} =\]
(256)0.16 × (256)0.09
When simplified \[(256) {}^{- ( 4^{- 3/2} )}\] is
If \[\frac{3^{5x} \times {81}^2 \times 6561}{3^{2x}} = 3^7\] then x =
If x = \[\sqrt[3]{2 + \sqrt{3}}\] , then \[x^3 + \frac{1}{x^3} =\]
Find:-
`32^(1/5)`
Simplify:
`(1^3 + 2^3 + 3^3)^(1/2)`
