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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`
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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`
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Simplify:-
`(1/3^3)^7`
