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
Simplify the following:
`(5xx25^(n+1)-25xx5^(2n))/(5xx5^(2n+3)-25^(n+1))`
Solve the following equation for x:
`2^(x+1)=4^(x-3)`
Solve the following equation for x:
`4^(2x)=1/32`
If 2x = 3y = 6-z, show that `1/x+1/y+1/z=0`
If `27^x=9/3^x,` find x.
Solve the following equation:
`8^(x+1)=16^(y+2)` and, `(1/2)^(3+x)=(1/4)^(3y)`
If `2^x xx3^yxx5^z=2160,` find x, y and z. Hence, compute the value of `3^x xx2^-yxx5^-z.`
Which one of the following is not equal to \[\left( \sqrt[3]{8} \right)^{- 1/2} ?\]
(256)0.16 × (256)0.09
The simplest rationalising factor of \[\sqrt[3]{500}\] is
