Harold Nelson
2/22/2022
A log function reverses the action of the exponential function with the same base.
This is one of two factual statements regarding this principle.
\[a^{log_a(x)} = x\]
\[log_aa^x = x\]
First, let’s look at a particular example. Consider the expression \[2^2*2^3\] This could be written as \[(2*2)*(2*2*2)\] And this in turn could be written as \[2^5\] The general principle is
\[a^m*a^n = a^{m+n}\]
\[log(a*b)=log(a)+log(b)\]
\[a^0 = 1\] Why would this have to be true? Consider the expression \[a^m*a^0\]
From Principle 2,
\[a^m*a^0 = a^{m+0} = a^m\] Therefore
\[a^0 = 1\]
\[log_a1=0\]
\[a^{-n} =\frac{1}{a^n}\] Why would this have to be true? From Principle 1,
\[a^n*a^{-n} = a^{n-n}=a^0=1\] So it is the reciprocal.
\[log_a(1/x^n) = -n* log_ax\]
This follows directly from Principle 4
\[a^m/a^n = a^{m-n}\]
\[log_a(m/n)=log_am-log_an\]
This is just a special case of Principle 4.
\[a^{-1} = 1/a\]
\[ log_a(1/b)=-log_ab\]
\[(a^m)^n = a^{m*n}\]
To see why this is true, consider a concrete example.
\[(2^2)^3 = (2*2)^3=(2*2)*(2*2)*(2*2)=2^6\]
\[log_a(b^n)=n*log_ab\]