Advertisements
Advertisements
Question
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
Solution
(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
RELATED QUESTIONS
Simplify the following
`(2x^-2y^3)^3`
Solve the following equations for x:
`3^(2x+4)+1=2.3^(x+2)`
Solve the following equation:
`4^(x-1)xx(0.5)^(3-2x)=(1/8)^x`
Solve the following equation:
`sqrt(a/b)=(b/a)^(1-2x),` where a and b are distinct primes.
Write the value of \[\left\{ 5( 8^{1/3} + {27}^{1/3} )^3 \right\}^{1/4} . \]
The value of x − yx-y when x = 2 and y = −2 is
The product of the square root of x with the cube root of x is
`(2/3)^x (3/2)^(2x)=81/16 `then x =
If a, m, n are positive ingegers, then \[\left\{ \sqrt[m]{\sqrt[n]{a}} \right\}^{mn}\] is equal to
If \[\sqrt{5^n} = 125\] then `5nsqrt64`=
