Advertisements
Advertisements
Question
If 49392 = a4b2c3, find the values of a, b and c, where a, b and c are different positive primes.
Advertisements
Solution
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
RELATED QUESTIONS
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrt(x^-3))^5`
Prove that:
`sqrt(1/4)+(0.01)^(-1/2)-(27)^(2/3)=3/2`
Show that:
`(x^(a^2+b^2)/x^(ab))^(a+b)(x^(b^2+c^2)/x^(bc))^(b+c)(x^(c^2+a^2)/x^(ac))^(a+c)=x^(2(a^3+b^3+c^3))`
Show that:
`(a^(x+1)/a^(y+1))^(x+y)(a^(y+2)/a^(z+2))^(y+z)(a^(z+3)/a^(x+3))^(z+x)=1`
If ax = by = cz and b2 = ac, show that `y=(2zx)/(z+x)`
Find the value of x in the following:
`(2^3)^4=(2^2)^x`
Find the value of x in the following:
`5^(x-2)xx3^(2x-3)=135`
If 102y = 25, then 10-y equals
If \[\frac{3^{2x - 8}}{225} = \frac{5^3}{5^x},\] then x =
\[\frac{1}{\sqrt{9} - \sqrt{8}}\] is equal to
