Question

Express \[{\log _2}(6!)\] in the form of \[a + {\log _2}b\] where \[a{\text{ and }}b\] are integers and \[b\] is the smallest possible value.

Answer 1

Hint: We will use the concepts of algebra related to exponents and logarithms.
Consider an equation \[{a^x} = b\].
Here, \[a\] is called the base and \[b\] is called the result. But there is another term \[x\] which is called the exponent. Exponent is the power to the base which defines the number of times the base is multiplied by itself.
Suppose, there is a term \[{3^5}\], which means that, 3 is multiplied by itself for five times.
So, \[{3^5} = 3 \times 3 \times 3 \times 3 \times 3\]
In the equation \[{a^x} = b\], the value of \[a\] can be found by making it the subject of the equation.
\[ \Rightarrow a = {\left( b \right)^{\dfrac{1}{x}}}\] or \[a = \sqrt[x]{b}\]
But, to find the value of \[x\], we need to apply a function called ‘Logarithm’.
So, it is defined as the reverse process of exponential function.
For equation \[{a^x} = b\], logarithm can be applied as \[x = {\log _a}b\]

Formula used:
There are few rules or limits for a logarithm. They are,
(1) Logarithm is not defined for a negative value i.e., \[\log n\] is not defined if \[n < 0\].
(2) Logarithm of 1 to any base is always 0. \[ \Rightarrow {\log _a}1 = 0\], because \[{a^0} = 1\] where \[a \in R\]
(3) Logarithm to base \[e = 2.71\] is represented as \[\ln b\] which is called natural logarithm.
There is a standard formula in logarithms which is \[{\log _n}(ab) = {\log _n}a + {\log _n}b\]
This is known as the logarithm rule of a product.

Complete step by step answer:
The given question is \[{\log _2}(6!)\], which can be written as,
\[ \Rightarrow {\log _2}(6!) = {\log _2}(6 \times 5 \times 4 \times 3 \times 2 \times 1)\]
We can rewrite some of these numbers as products of two other numbers.
\[ \Rightarrow {\log _2}(6 \times 5 \times 4 \times 3 \times 2 \times 1) = {\log _2}((2 \times 3) \times 5 \times (2 \times 2) \times 3 \times 2 \times 1)\]
Now, on groping all the 2’s, we get,
\[ \Rightarrow {\log _2}(6 \times 5 \times 4 \times 3 \times 2 \times 1) = {\log _2}({2^4} \times ({3^2} \times 5))\]
Now, according to product rule, we can write this as,
\[ \Rightarrow {\log _2}({2^4} \times ({3^2} \times 5)) = {\log _2}{2^4} + {\log _2}({3^2} \times 5)\]
Now, we know the formula, \[{\log _a}{m^n} = n{\log _a}m\]
So, applying this, we get,
\[ \Rightarrow 4{\log _2}2 + {\log _2}({3^2} \times 5)\]
And here, we apply \[{\log _a}a = 1\]
\[ \Rightarrow 4(1) + {\log _2}(45)\]
\[ \Rightarrow 4 + {\log _2}(45)\]
So, we have expressed the logarithm \[{\log _2}(6!)\] in the form of \[a + {\log _2}(b)\], where \[a = 4\] & \[b = 45\].

Note:
Also remember another identity which is \[{\log _n}\left( {\dfrac{a}{b}} \right) = {\log _n}a - {\log _n}b\]
Logarithm to base 10 is represented as just \[\log a\] (without representing any base)
Also remember a rule for \[a > 1\] and \[x > y\], \[{\log _a}x > {\log _a}y\].
For \[0 < a < 1\] and \[x > y\], \[{\log _a}x < {\log _a}y\].
These are some inequalities of logarithms.