Advertisements
Advertisements
प्रश्न
Find the value of x in the following:
`2^(5x)div2x=root5(2^20)`
Advertisements
उत्तर
Given `2^(5x)div2x=root5(2^20)`
By using rational exponents `a^m/a^n=a^(m-n)` we get,
`2^(5x-x)=2^(20xx1/5)`
`2^(5x-x)=2^4`
On equating the exponents we get,
5x - x = 4
4x = 4
x = 4/4
x = 1
The value of x = 1
APPEARS IN
संबंधित प्रश्न
If a = 3 and b = -2, find the values of :
ab + ba
Simplify:
`((25)^(3/2)xx(243)^(3/5))/((16)^(5/4)xx(8)^(4/3))`
Prove that:
`((0.6)^0-(0.1)^-1)/((3/8)^-1(3/2)^3+((-1)/3)^-1)=(-3)/2`
If 2x = 3y = 12z, show that `1/z=1/y+2/x`
If 3x = 5y = (75)z, show that `z=(xy)/(2x+y)`
Find the value of x in the following:
`(sqrt(3/5))^(x+1)=125/27`
Determine `(8x)^x,`If `9^(x+2)=240+9^x`
If `3^(4x) = (81)^-1` and `10^(1/y)=0.0001,` find the value of ` 2^(-x+4y)`.
The value of \[\sqrt{5 + 2\sqrt{6}}\] is
Simplify:-
`(1/3^3)^7`
