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Chapter 7: Problem 13

Explain how it follows from the definition of logarithm that a. \(\log _{b}\left(b^{x}\right)=x\), for all real numbers \(x\). b. \(b^{\log _{b} x}=x\), for all positive real numbers \(x\).

Short Answer

Expert verified
In short: a. For \( \log _{b}\left(b^{x}\right)=x \), according to the definition of logarithm, we rewrite it as \( b^{x} = b^{\log _{b}\left(b^{x}\right)} \). Since both sides equal to \(b^x\), it implies that \( \log _{b}\left(b^{x}\right)=x \) for all real numbers \(x\). b. For \( b^{\log _{b} x}=x \), we rewrite it using the definition of logarithm to get \( \log _{b}\left(b^{\log_{b} x}\right) = \log_{b}(x) \). Using the result from Statement A, we have \( b^{\log _{b} x}=x \) for all positive real numbers \(x\). Both statements follow from the definition of logarithm.

Step by step solution

01

Statement A: Understanding the definition of logarithm

The definition of a logarithm is as follows: \[\log_b{y} = x\] if and only if \[b^x = y\]. In this case, we are given \(\log _{b}\left(b^{x}\right)=x\). Now we can apply this definition to the equation.
02

Statement A: Rewriting the statement using the definition of logarithm

According to the definition, we can rewrite the statement as: \[b^{x} = b^{\log _{b}\left(b^{x}\right)}\]
03

Statement A: Proving the equality

The left side of the equation is simply \(b^{x}\), and if we look at the right side, \(b^{\log _{b} \left(b^{x}\right)}\), this equals to \(b^{x}\) as well by definition of logarithm. Thus, we have: \[b^{x} = b^{\log _{b} \left(b^{x}\right)}\] which implies that \[\log _{b}\left(b^{x}\right)=x\] for all real numbers \(x\). Now, let's move on to Statement B.
04

Statement B: Understanding the definition of logarithm

As before, we will use the definition of logarithm: \[\log_b{y} = x\] if and only if \[b^x = y\]. In this case, we are given \(b^{\log _{b} x}=x\). Now we can apply this definition to the equation.
05

Statement B: Rewriting the statement using the definition of logarithm

According to the definition, we can rewrite the statement as: \[\log _{b}\left(b^{\log_{b} x}\right) = \log_{b}(x)\]
06

Statement B: Proving the equality

We have already proven Statement A as \[\log _{b}\left(b^{x}\right)=x\]. Using this result, we can say that: \[\log _{b}\left(b^{\log_{b} x}\right) = \log_{b}(x)\] Which implies that: \[b^{\log _{b} x}=x\] for all positive real numbers \(x\). In conclusion, we've shown that both statements follow from the definition of logarithm.

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Most popular questions from this chapter

Use the definition of logarithm to prove that for any positive real number \(b\) with \(b \neq 1, \log _{b} b=1\).

Let \(A\) be a set of six positive integers each of which is less than 13. Show that there must be two distinct subsets of \(A\) whose elements when added up give the same sum. (For example, if \(A=\\{5,12,10,1,3,4\\}\), then the elements of the subsets \(S_{1}=\\{1,4,10\\}\) and \(S_{2}=\\{5,10\\}\) both add up to 15.)

Let \(A=\\{1,2,3,4,5\\}\) and define a function \(F: \mathscr{P}(A) \rightarrow \mathbf{Z}\) as follows: For all sets \(X\) in \(\mathscr{P}(A)\), \(F(X)=\) \(= \begin{cases}0 & \text { if } X \text { has an even } \\ & \text { number of elements } \\ 1 & \text { if } X \text { has an odd } \\ & \text { number of elements }\end{cases}\) Find the following a. \(F(\\{1,3,4\\})\) b. \(F(b)\) c. \(F(\\{2,3\\})\) d. \(F(\\{2,3,4,5\\})\)

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