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A number $x$ factors as $107\cdot109^5$, and $107$ and $109$ are primes. What is the exponent of $109$ in the prime factorization of $x^{11}$?

Respuesta :

Answer:

*55*

Yeah it s very easy Bro!

Step-by-step explanation:

Watch we have something like 109^5=x

then x^11=109^x

11*5=55

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Answer:

55

Step-by-step explanation:

We have x^11 = (107 x 109^5)^11 = 107^11 (109^5)^11 = 107^11 109^55, so our answer is 55.

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