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
Simplify the following
`((4xx10^7)(6xx10^-5))/(8xx10^4)`
If a = 3 and b = -2, find the values of :
(a + b)ab
Simplify the following:
`(5^(n+3)-6xx5^(n+1))/(9xx5^x-2^2xx5^n)`
Simplify:
`((5^-1xx7^2)/(5^2xx7^-4))^(7/2)xx((5^-2xx7^3)/(5^3xx7^-5))^(-5/2)`
Show that:
`[{x^(a(a-b))/x^(a(a+b))}div{x^(b(b-a))/x^(b(b+a))}]^(a+b)=1`
Solve the following equation:
`3^(x+1)=27xx3^4`
Write the value of \[\sqrt[3]{7} \times \sqrt[3]{49} .\]
If \[\sqrt{5^n} = 125\] then `5nsqrt64`=
\[\frac{5^{n + 2} - 6 \times 5^{n + 1}}{13 \times 5^n - 2 \times 5^{n + 1}}\] is equal to
If \[x = \sqrt{6} + \sqrt{5}\],then \[x^2 + \frac{1}{x^2} - 2 =\]
