Advertisements
Advertisements
प्रश्न
If 49392 = a4b2c3, find the values of a, b and c, where a, b and c are different positive primes.
Advertisements
उत्तर
First find out the prime factorisation of 49392.

It can be observed that 49392 can be written as `2^4xx3^2xx7^3,` where 2, 3 and 7 are positive primes.
`therefore49392=2^4 3^2 7^3=a^4b^2c^3`
⇒ a = 2, b = 3, c = 7
APPEARS IN
संबंधित प्रश्न
Simplify the following
`((4xx10^7)(6xx10^-5))/(8xx10^4)`
Simplify the following:
`(3^nxx9^(n+1))/(3^(n-1)xx9^(n-1))`
Solve the following equations for x:
`3^(2x+4)+1=2.3^(x+2)`
If `a=xy^(p-1), b=xy^(q-1)` and `c=xy^(r-1),` prove that `a^(q-r)b^(r-p)c^(p-q)=1`
If 2x = 3y = 12z, show that `1/z=1/y+2/x`
Solve the following equation:
`3^(x-1)xx5^(2y-3)=225`
If a and b are different positive primes such that
`((a^-1b^2)/(a^2b^-4))^7div((a^3b^-5)/(a^-2b^3))=a^xb^y,` find x and y.
If 3x-1 = 9 and 4y+2 = 64, what is the value of \[\frac{x}{y}\] ?
If 102y = 25, then 10-y equals
If \[\sqrt{2} = 1 . 4142\] then \[\sqrt{\frac{\sqrt{2} - 1}{\sqrt{2} + 1}}\] is equal to
