Advertisements
Advertisements
Question
If a and b are distinct primes such that `root3 (a^6b^-4)=a^xb^(2y),` find x and y.
Advertisements
Solution
Given `root3 (a^6b^-4)=a^xb^(2y)`
`rArr(a^6b^-4)^(1/3)=a^xb^(2y)`
`rArra^(6xx1/3)b^(-4xx1/3)=a^xb^(2y)`
`rArra^2b^(-4/3)=a^xb^(2y)`
⇒ x = 2 and y = -2/3
APPEARS IN
RELATED QUESTIONS
Find:-
`9^(3/2)`
Prove that:
`sqrt(1/4)+(0.01)^(-1/2)-(27)^(2/3)=3/2`
Prove that:
`(3^-3xx6^2xxsqrt98)/(5^2xxroot3(1/25)xx(15)^(-4/3)xx3^(1/3))=28sqrt2`
If ax = by = cz and b2 = ac, show that `y=(2zx)/(z+x)`
Find the value of x in the following:
`(root3 4)^(2x+1/2)=1/32`
Find the value of x in the following:
`(sqrt(3/5))^(x+1)=125/27`
If (23)2 = 4x, then 3x =
If \[\frac{2^{m + n}}{2^{n - m}} = 16\], \[\frac{3^p}{3^n} = 81\] and \[a = 2^{1/10}\],than \[\frac{a^{2m + n - p}}{( a^{m - 2n + 2p} )^{- 1}} =\]
Simplify:-
`(1/3^3)^7`
Simplify:
`11^(1/2)/11^(1/4)`
