Question

The total number of prime factors of (9 x 2)24 (4 x 8)15​

Answer 1

Answer:

The prime factorisation of 24 gives 2 × 2 × 2 × 3 = 23 × 3

Answer 2

The total number of prime factors of $(9\times 2)^{24}\times (4\times 8)^{15}$ is 147.

Step-by-step explanation:

First term

$(9\times 2)^{24}$

$=(3^{2}\times 2)^{24}$

$=3^{2\times 24}\times 2^{24}$

• since $(a\times b)^{m}=a^{m}\times b^{m}$

$=3^{48}\times 2^{24}$

So, there are (48 + 24) = 72 prime factors in $(9\times 2)^{24}$.

Second term

$(4\times 8)^{15}$

$=(2^{2}\times 2^{3})^{15}$

$=(2^{2+3})^{15}$

$=(2^{5})^{15}$

$=2^{75}$

So, there are 75 prime factors in $(4\times 8)^{15}$.

Thus the total number of prime factors in $(9\times 2)^{24}\times (4\times 8)^{15}$ is (72 + 75) = 147.

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