Answer
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Hint: First solve this in prime factorization method and then convert into prime factors of exponential form.
Starting with the first prime number we can write $270$as
$270 = 2 \times 135$
Then converting $135$ into the factors of next prime number $3$we have
$270 = 2 \times 3 \times 45$
Again converting $45$into the prime factors of $3$ we have
$270 = 2 \times 3 \times 3 \times 15$
Now at last converting $15$into the prime factors of $3$and $5$ we have
$270 = 2 \times 3 \times 3 \times 3 \times 5$
Finally, converting them into exponential form we have
\[270 = {2^1} \times {3^3} \times {5^1}\]
Thus, the product of prime factors in the exponential form of \[270\]is \[{2^1} \times {3^3} \times{5^1}\]
Note: In this type of problems first write prime factors of lower prime numbers and then to higher
prime numbers.
Starting with the first prime number we can write $270$as
$270 = 2 \times 135$
Then converting $135$ into the factors of next prime number $3$we have
$270 = 2 \times 3 \times 45$
Again converting $45$into the prime factors of $3$ we have
$270 = 2 \times 3 \times 3 \times 15$
Now at last converting $15$into the prime factors of $3$and $5$ we have
$270 = 2 \times 3 \times 3 \times 3 \times 5$
Finally, converting them into exponential form we have
\[270 = {2^1} \times {3^3} \times {5^1}\]
Thus, the product of prime factors in the exponential form of \[270\]is \[{2^1} \times {3^3} \times{5^1}\]
Note: In this type of problems first write prime factors of lower prime numbers and then to higher
prime numbers.
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