Advertisements
Advertisements
प्रश्न
Prove that:
`(2^n+2^(n-1))/(2^(n+1)-2^n)=3/2`
Advertisements
उत्तर
We have to prove that `(2^n+2^(n-1))/(2^(n+1)-2^n)=3/2`
Let x = `(2^n+2^(n-1))/(2^(n+1)-2^n)`
`=(2^n(1+1xx2^-1))/(2^n(2^1-1))`
`=(1+1/2)/(2-1)`
`rArrx=3/2`
Hence, `(2^n+2^(n-1))/(2^(n+1)-2^n)=3/2`
APPEARS IN
संबंधित प्रश्न
Solve the following equations for x:
`3^(2x+4)+1=2.3^(x+2)`
Given `4725=3^a5^b7^c,` find
(i) the integral values of a, b and c
(ii) the value of `2^-a3^b7^c`
Show that:
`(x^(1/(a-b)))^(1/(a-c))(x^(1/(b-c)))^(1/(b-a))(x^(1/(c-a)))^(1/(c-b))=1`
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:
`(sqrt(3/5))^(x+1)=125/27`
When simplified \[(256) {}^{- ( 4^{- 3/2} )}\] is
If \[2^{- m} \times \frac{1}{2^m} = \frac{1}{4},\] then \[\frac{1}{14}\left\{ ( 4^m )^{1/2} + \left( \frac{1}{5^m} \right)^{- 1} \right\}\] is equal to
If \[\sqrt{2^n} = 1024,\] then \[{3^2}^\left( \frac{n}{4} - 4 \right) =\]
\[\frac{1}{\sqrt{9} - \sqrt{8}}\] is equal to
