Advertisements
Advertisements
प्रश्न
If a and b are distinct primes such that `root3 (a^6b^-4)=a^xb^(2y),` find x and y.
Advertisements
उत्तर
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
संबंधित प्रश्न
Simplify:
`(0.001)^(1/3)`
Simplify:
`((5^-1xx7^2)/(5^2xx7^-4))^(7/2)xx((5^-2xx7^3)/(5^3xx7^-5))^(-5/2)`
Prove that:
`(2^n+2^(n-1))/(2^(n+1)-2^n)=3/2`
If `3^(4x) = (81)^-1` and `10^(1/y)=0.0001,` find the value of ` 2^(-x+4y)`.
If 24 × 42 =16x, then find the value of x.
Which of the following is (are) not equal to \[\left\{ \left( \frac{5}{6} \right)^{1/5} \right\}^{- 1/6}\] ?
If (23)2 = 4x, then 3x =
The value of \[\left\{ \left( 23 + 2^2 \right)^{2/3} + (140 - 19 )^{1/2} \right\}^2 ,\] is
If \[x = \sqrt{6} + \sqrt{5}\],then \[x^2 + \frac{1}{x^2} - 2 =\]
If \[\sqrt{13 - a\sqrt{10}} = \sqrt{8} + \sqrt{5}, \text { then a } =\]
