Problem 90 - College Algebra

Question

Explain why the logarithm of 1 with base \(b\) is 0.

Step-by-Step Solution

Answer

The logarithm of 1 with any base \(b\) is 0, denoted as \(\log_b(1) = 0\), as according to the properties of exponentiation any number raised to the power 0 gives 1.

Step 1: Understand the logarithm function

The logarithm function is the inverse of the exponential function. Given a logarithm function with base \(b\) and argument \(x\), denoted as \(\log_b(x)\), it represents the exponent to which the base \(b\) must be raised to get \(x\). In other words, if we have \(b^{y} = x\), then \(\log_b(x) = y\), where \(b > 0\), \(b \neq 1\), and \(x > 0\).

Step 2: Apply the logarithm property

We can apply this property to the problem at hand. We are asked to find \(\log_b(1)\) for an arbitrary base \(b\). So we need to find some number \(y\) such that raising \(b\) to the power of \(y\) gives us \(1\).

Step 3: Reason about the exponent needed

It's a common mathematical fact that any number to the power of 0 is 1. That is, \(b^{0} = 1\) for any number \(b\). This is because any number multiplied by itself zero times results in 1. So, to get an argument of 1 for our logarithm function, we need an exponent of 0.

Step 4: Input into the logarithm function

So based on Step 3, we can say that \(\log_b(1) = 0\). This is because \(b^{0} = 1\), which is a property of exponentiation. Therefore any logarithm having an argument of 1 is 0, no matter what the base is.