We note that by definition,
\[{b^{{{\log }_b}N}} = N\]
For example,
\[\begin{align}&\log_28=3,\qquad\qquad \quad2^{\log_28}=2^3=8\\&\log_2\left(\frac1{16}\right)=4,\qquad2^{\log_2\left(\frac1{16}\right)}=2^{4}=\frac1{16}\end{align}\]
Also, negative bases are not allowed in logarithms. We discussed this in the section on Exponents: if the base is negative, then all powers of the base will not be real numbers. Thus, for the term \(log_bN\) to be welldefined, we must have \( b>0\).
Next, we note that for \(log_bN\) to be a real number, N must be a positive number. Why is this so? Suppose that N is negative. Then, \(log_bN\) is a number e such that \(b^e=N\). But if e is a real number \(b^e\) , will always be positive (we discussed this: exponential terms always have positive values). Thus, \(b^e=N\) is not possible if N is negative.
We also note that b cannot be equal to 1. Let us understand why. Suppose that b=1. Let N be any number. Think of the term \(log_1N\) . This is a number e such that \(1^e=N\). But 1 raised to any power is 1. Thus, we see that there can be no value possible for \(log_1N\), since no matter what power you raise 1 to, you’ll never obtain N. This means that 1 is not a valid base for logs.
Finally, we note that \({\log _b}1 = 0\) for any base b. This is because any number raised to power 0 is 1, so that
\[b^0=1\;\;\Rightarrow\;\;\log_b1=0\]
We summarize all these properties below:

The term \(log_bN\) is well defined if
a. The base b is positive, and not equal to 1.
b. The number N is positive.

The log of 1 to any base is 0.