Advertisements
Advertisements
प्रश्न
Find the value of x in the following:
`(root3 4)^(2x+1/2)=1/32`
Advertisements
उत्तर
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
संबंधित प्रश्न
If a = 3 and b = -2, find the values of :
ab + ba
Solve the following equation for x:
`4^(2x)=1/32`
Assuming that x, y, z are positive real numbers, simplify the following:
`root5(243x^10y^5z^10)`
Simplify:
`(sqrt2/5)^8div(sqrt2/5)^13`
Simplify:
`((5^-1xx7^2)/(5^2xx7^-4))^(7/2)xx((5^-2xx7^3)/(5^3xx7^-5))^(-5/2)`
\[\frac{5^{n + 2} - 6 \times 5^{n + 1}}{13 \times 5^n - 2 \times 5^{n + 1}}\] is equal to
The value of \[\frac{\sqrt{48} + \sqrt{32}}{\sqrt{27} + \sqrt{18}}\] is
The value of \[\sqrt{5 + 2\sqrt{6}}\] is
If \[\sqrt{2} = 1 . 414,\] then the value of \[\sqrt{6} - \sqrt{3}\] upto three places of decimal is
If \[\sqrt{13 - a\sqrt{10}} = \sqrt{8} + \sqrt{5}, \text { then a } =\]
