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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 well-defined, 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:

1. 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.

2. The log of 1 to any base is 0.

Exponents and Logarithms
Exponents and Logarithms
Grade 9 | Questions Set 2
Exponents and Logarithms