Advertisements
Advertisements
Question
If 1176 = `2^axx3^bxx7^c,` find the values of a, b and c. Hence, compute the value of `2^axx3^bxx7^-c` as a fraction.
Advertisements
Solution
First find the prime factorisation of 1176.
It can be observed that 1176 can be written as `2^3xx3^1xx7^2`
`1176=2^a3^b7^c=2^3 3^1 7^2`
So, a = 3, b = 1 and c = 2.
Therefore, the value of `2^a xx3^bxx76-c` is `2^3xx3^1xx7^-2=8xx3xx1/49=24/49`
APPEARS IN
RELATED QUESTIONS
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrt(x^-3))^5`
If 2x = 3y = 12z, show that `1/z=1/y+2/x`
Find the value of x in the following:
`2^(5x)div2x=root5(2^20)`
Find the value of x in the following:
`(sqrt(3/5))^(x+1)=125/27`
Write \[\left( \frac{1}{9} \right)^{- 1/2} \times (64 )^{- 1/3}\] as a rational number.
The value of \[\left\{ 2 - 3 (2 - 3 )^3 \right\}^3\] is
The value of \[\left\{ 8^{- 4/3} \div 2^{- 2} \right\}^{1/2}\] is
If \[\frac{3^{5x} \times {81}^2 \times 6561}{3^{2x}} = 3^7\] then x =
The simplest rationalising factor of \[\sqrt[3]{500}\] is
The value of \[\frac{\sqrt{48} + \sqrt{32}}{\sqrt{27} + \sqrt{18}}\] is
