Advertisements
Advertisements
प्रश्न
If \[4x - 4 x^{- 1} = 24,\] then (2x)x equals
विकल्प
\[5\sqrt{5}\]
\[\sqrt{5}\]
\[25\sqrt{5}\]
125
उत्तर
We have to find the value of `(2x)^x`if `4^x - 4^(x-1) = 24`
So,
Taking 4x as common factor we get
`4^x (1- 1/4) = 24`
`4^x (1-4^-1) = 24`
`4^x ((1xx4)/(1 xx4)-1/4) = 24`
`4^4 ((4-1)/4)= 24`
`4^x xx 3/4 = 24`
`4^x = 24 xx 4/3`
`4^x = 32`
`2^(2x) =2^5`
By equating powers of exponents we get
`2x = 5 `
`x=5/2`
By substituting `x=5/2` in `(2x)^x` we get
`(2x)^x=(2xx 5/2)^(5/2)`
= `(2xx5/2)^(5/2)`
`=5^(5/2)`
`=5^(5 xx1/2)`
`(2x)^x = 2sqrt(5^5)`
`=2sqrt (5xx5xx5xx5xx5)`
`= 5xx5 2sqrt5`
= `25sqrt5`
APPEARS IN
संबंधित प्रश्न
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrt(x^-3))^5`
If `27^x=9/3^x,` find x.
Find the value of x in the following:
`(2^3)^4=(2^2)^x`
Find the value of x in the following:
`(3/5)^x(5/3)^(2x)=125/27`
If `x = a^(m+n),` `y=a^(n+l)` and `z=a^(l+m),` prove that `x^my^nz^l=x^ny^lz^m`
Write the value of \[\left\{ 5( 8^{1/3} + {27}^{1/3} )^3 \right\}^{1/4} . \]
The seventh root of x divided by the eighth root of x is
If a, b, c are positive real numbers, then \[\sqrt{a^{- 1} b} \times \sqrt{b^{- 1} c} \times \sqrt{c^{- 1} a}\] is equal to
If \[\frac{3^{2x - 8}}{225} = \frac{5^3}{5^x},\] then x =
If \[\sqrt{2} = 1 . 4142\] then \[\sqrt{\frac{\sqrt{2} - 1}{\sqrt{2} + 1}}\] is equal to
