Express log2(6!) in form a + log2(b) where a and b are integers and b is the smallest possible value?

Question

Express #log_2(6!)# in the form #a + log_2(b)#, where a, b #in# z, and b is the smallest possible value

Anna Allen · Accepted Answer

#log_2(6!)==4+log_2(45)# #log_2(6!)=log_2(2*3*4*5*6)# #=log_2(2*4)+log_2(3*5*2*3)# #=log_2(2*4)+log_2(2)+log_2(3*5*3)# #=log_2(2*4*2)+log_2(3*5*3)# #=log_2(16)+log_2(45)# #=cancel(log_2(2^4))^(color(red)(=4))+log_2(45)# #=4+log_2(45)# Because #gcd(45,2)=1#, #45# is the smallest #b# you can have. \0/ here's our answer !

Greyson Brown · Answer

undefined 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:

1. Calculate $ \log_2(6!) $.
2. Rewrite the result in the desired form.

1. Calculate $ \log_2(6!) $:

$$ 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 $$
$$ 6! = 720 $$

$$ \log_2(720) $$

2. 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 $.

$$ \log_2(6!) = 10 + \log_2(5) $$