Advertisements
Advertisements
Question
If `1176=2^a3^b7^c,` find a, b and c.
Advertisements
Solution
First find out the prime factorisation of 1176.

It can be observed that 1176 can be written as `2^3xx3^1xx7^2.`
`1176=2^3 3^1 7^2 = 2^a3^b7^c`
Hence, a = 3, b = 1 and c = 2.
APPEARS IN
RELATED QUESTIONS
Solve the following equation for x:
`2^(3x-7)=256`
If 49392 = a4b2c3, find the values of a, b and c, where a, b and c are different positive primes.
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrt(x^-3))^5`
Simplify:
`(sqrt2/5)^8div(sqrt2/5)^13`
Simplify:
`((5^-1xx7^2)/(5^2xx7^-4))^(7/2)xx((5^-2xx7^3)/(5^3xx7^-5))^(-5/2)`
Show that:
`(x^(1/(a-b)))^(1/(a-c))(x^(1/(b-c)))^(1/(b-a))(x^(1/(c-a)))^(1/(c-b))=1`
If `27^x=9/3^x,` find x.
Simplify:
`root(lm)(x^l/x^m)xxroot(mn)(x^m/x^n)xxroot(nl)(x^n/x^l)`
If (x − 1)3 = 8, What is the value of (x + 1)2 ?
The seventh root of x divided by the eighth root of x is
