Express log2(6!) in form a + log2(b) where a and b are integers and b is the smallest possible value?
Express #log_2(6!)# in the form #a + log_2(b)# , where a, b #in# z, and b is the smallest possible value
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To express ( \log_2(6!) ) in the form ( a + \log_2(b) ) where ( a ) and ( b ) are integers and ( b ) is the smallest possible value, follow these steps:

Calculate ( \log_2(6!) ).

Rewrite the result in the desired form.

Calculate ( \log_2(6!) ):
[ 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 ] [ 6! = 720 ]
[ \log_2(720) ]
 Break down 720 into factors of powers of 2 to simplify the logarithm:
[ 720 = 2^6 \times 45 ] [ 720 = 2^6 \times 3^2 \times 5 ]
Using the properties of logarithms:
[ \log_2(720) = \log_2(2^6 \times 3^2 \times 5) ] [ \log_2(720) = \log_2(2^6) + \log_2(3^2) + \log_2(5) ]
Apply the power rule of logarithms:
[ \log_2(2^6) = 6 ] [ \log_2(3^2) = 2 \times 2 = 4 ]
Combine the results:
[ \log_2(720) = 6 + 4 + \log_2(5) ]
[ \log_2(720) = 10 + \log_2(5) ]
So, ( a = 10 ) and ( b = 5 ).
[ \log_2(6!) = 10 + \log_2(5) ]
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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