Question

Evaluate \[{\log _2}^{64}\].

Answer 1

Hint: Express \[64\] as a power of \[2\], i.e. \[{2^n}\], then use the properties of logarithm. when nothing is written in the base of log we should understand that it’s ‘e’ but in this case it’s given to us as 2.

Complete step by step solution:
Given: \[{\log _2}^{64}\], express \[64\] in the form \[{2^n}\] .
\[64 = {2^6}\]
\[\therefore {\log _2}^{64} = {\log _2}^{{2^6}}\]
From properties of logarithm:
\[{\log _b}^{{a^n}} = n{\log _b}^a\]
\[\therefore {\log _2}^{{2^6}} = 6{\log _2}^2\]
Again \[{\log _a}^a = 1\]
\[\therefore {\log _2}^2 = 1\]
\[\therefore 6{\log _2}^2 = 6\]

Hence,
\[{\log _2}^{64}\] \[ = \] \[6\].

Additional Information:
A logarithm is the inverse function of exponentiation. That means the logarithm of a given number \[x\] is the exponent to which another fixed number, the base \[x\], must be raised, to produce that number \[x\].
\[{\log _a}^{\left( x \right)} = y\] exactly if \[{a^y} = x\] and \[x > 0\], \[a > 0\], \[a \ne 1\]

Note:
Students must memorise the following properties of logarithm for solving problems like this:
1. \[{\log _a}^{\left( {xy} \right)} = {\log _a}^x + {\log _a}^y\]
2. \[{\log _a}^{\left( {\dfrac{x}{y}} \right)} = {\log _a}^x - {\log _a}^y\]
3. \[{\log _a}^{{{\left( x \right)}^y}} = y{\log _a}^x\]
4. \[{\log _a}^x = \dfrac{1}{{{{\log }_x}^a}}\]
5. \[{\log _a}^x = \dfrac{{{{\log }_b}^x}}{{{{\log }_b}^a}}\]
6. \[{\log _a}^a = 1\]
7. \[{\log _a}^{\left( 0 \right)} = \] undefined
8. \[{\log _a}^{\left( 1 \right)} = \] \[0\]