Advertisements
Advertisements
प्रश्न
Given `4725=3^a5^b7^c,` find
(i) the integral values of a, b and c
(ii) the value of `2^-a3^b7^c`
Advertisements
उत्तर
(i) Given `4725=3^a5^b7^c`
First find out the prime factorisation of 4725.
It can be observed that 4725 can be written as `3^3xx5^2xx7^1.`
`therefore4725 = 3^a5^b7^c=3^3 5^2 7^1`
Hence, a = 3, b = 2 and c = 1.
(ii)
When a = 3, b = 2 and c = 1,
`2^-a3^b7^c`
`=2^-3xx3^2xx7^1`
`=1/8xx9xx7`
`=63/8`
APPEARS IN
संबंधित प्रश्न
Assuming that x, y, z are positive real numbers, simplify the following:
`root5(243x^10y^5z^10)`
Simplify:
`(0.001)^(1/3)`
Show that:
`1/(1+x^(a-b))+1/(1+x^(b-a))=1`
Find the value of x in the following:
`(root3 4)^(2x+1/2)=1/32`
Solve the following equation:
`3^(x+1)=27xx3^4`
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 `x = a^(m + n), y = a^(n + l)` and `z = a^(l + m),` prove that `x^my^nz^l = x^ny^lz^m`
When simplified \[\left( - \frac{1}{27} \right)^{- 2/3}\] is
If a, m, n are positive ingegers, then \[\left\{ \sqrt[m]{\sqrt[n]{a}} \right\}^{mn}\] is equal to
The simplest rationalising factor of \[\sqrt{3} + \sqrt{5}\] is ______.
