Advertisements
Advertisements
प्रश्न
Solve the following equation:
`sqrt(a/b)=(b/a)^(1-2x),` where a and b are distinct primes.
Advertisements
उत्तर
`sqrt(a/b)=(b/a)^(1-2x)`
`rArr(a/b)^(1/2)=(a/b)^(-(1-2x))`
`rArr1/2=-(1-2x)`
`rArr1/2=2x - 1`
`rArr1/2+1=2x`
`rArr1/2+(1xx2)/(1xx2)=2x`
`rArr1/2+2/2=2x`
`rArr(1+2)/2=2x`
`rArr3/2=2x`
`rArrx=3/4`
APPEARS IN
संबंधित प्रश्न
Simplify the following
`(a^(3n-9))^6/(a^(2n-4))`
Prove that:
`1/(1 + x^(b - a) + x^(c - a)) + 1/(1 + x^(a - b) + x^(c - b)) + 1/(1 + x^(b - c) + x^(a - c)) = 1`
Assuming that x, y, z are positive real numbers, simplify the following:
`(x^-4/y^-10)^(5/4)`
If `3^(x+1)=9^(x-2),` find the value of `2^(1+x)`
Solve the following equation:
`3^(x-1)xx5^(2y-3)=225`
If `x = a^(m + n), y = a^(n + l)` and `z = a^(l + m),` prove that `x^my^nz^l = x^ny^lz^m`
The square root of 64 divided by the cube root of 64 is
If \[\frac{x}{x^{1 . 5}} = 8 x^{- 1}\] and x > 0, then x =
If \[\frac{2^{m + n}}{2^{n - m}} = 16\], \[\frac{3^p}{3^n} = 81\] and \[a = 2^{1/10}\],than \[\frac{a^{2m + n - p}}{( a^{m - 2n + 2p} )^{- 1}} =\]
Which of the following is equal to x?
