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Chapter 4: Problem 122

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

Short Answer

Expert verified
The logarithm of 1 with any base \(b\) is 0 because, by definition, the logarithm of a number is the exponent to which the base must be raised to obtain that number. In this case, any number \(b\) raised to the 0 power equals 1.

Step by step solution

01

Understanding Logarithms

The logarithm, \( \log_b(a) \), is defined by the equation \( b^{log_b(a)} = a \). In other words, the logarithm base \( b \) of \( a \) is the exponent to which you have to raise \( b \) to obtain \( a \).
02

Applying the Logarithm Definition

In this case, the base (\(b\)) doesn't matter, and the question is asking why \( \log_b(1) \) equals 0. Given the logarithm definition, we therefore have \( b^x = 1 \).
03

Solving the Equation

The only way for any non-zero number \( b \) raised to an exponent to become 1 is when the exponent is 0. Thus, \( b^0 = 1 \).
04

Concluding

Therefore, \( \log_b(1) = 0 \) since 0 is the value we have to raise the base \( b \) to in order to get 1.

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Most popular questions from this chapter

In Exercises \(53-56,\) rewrite the equation in terms of base \(e\). Express the answer in terms of a natural logarithm and then round to three decimal places. $$ y=4.5(0.6)^{x} $$

In parts (a)-(c), graph \(f\) and \(g\) in the same viewing rectangle. a. \(f(x)=\ln (3 x), g(x)=\ln 3+\ln x\) b. \(f(x)=\log \left(5 x^{2}\right), g(x)=\log 5+\log x^{2}\) c. \(f(x)=\ln \left(2 x^{3}\right), g(x)=\ln 2+\ln x^{3}\) d. Describe what you observe in parts (a)-(c). Generalize this observation by writing an equivalent expression for \(\log _{b}(M N),\) where \(M>0\) and \(N>0\) e. Complete this statement: The logarithm of a product is equal to _________.

Determine whether each statement makes sense or does not make sense, and explain your reasoning. Because the equations \(2^{x}=15\) and \(2^{x}=16\) are similar, I solved them using the same method.

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