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Chapter 8: Problem 16

If \(a>0, a \neq 1\), and \(x\) and \(y\) belong to \((0, \infty)\), prove that \(\log _{a}(x y)=\log _{a} x+\log _{a} y\).

Short Answer

Expert verified
The proof shows that the logarithm of the product of two numbers is equal to the sum of the logarithms of these numbers in the same base, given the conditions specified: \(\log_{a}(xy) = \log_{a}x + \log_{a}y\).

Step by step solution

01

Applying the Definition of Logarithms

Start by noting that by definition, if \(a^u = x\) then \(\log_{a}{x} = u\). Writing \(\log_{a}{x}\) and \(\log_{a}{y}\) in this exponents form, we get \(a^{\log_{a}{x}} = x\) and \(a^{\log_{a}{y}} = y\).
02

Multiplying the equations

Next, multiply these two equations together. The multiplication gives us \(a^{\log_{a}{x}} \cdot a^{\log_{a}{y}} = x \cdot y\). According to rules of exponent, when we multiply two terms with the same base, we can add the exponents. Therefore, the left side of the equation becomes \(a^{\log_{a}{x} + \log_{a}{y}}\).
03

Final simplification

So now we have \(a^{\log_{a}{x} + \log_{a}{y}} = x \cdot y\). Using the rules from Step 1 again (but backwards), we obtain \(\log_{a}{x \cdot y} = \log_{a}{x} + \log_{a}{y}\), which is what we wanted to prove.

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Most popular questions from this chapter

Suppose the sequence \(\left(f_{n}\right)\) converges uniformly to \(f\) on the set \(A\), and suppose that each \(f_{n}\) is bounded on \(A\). (That is, for each \(n\) there is a constant \(M_{n}\) such that \(\left|f_{n}(x)\right| \leq M_{n}\) for all \(x \in A\).) Show that the function \(f\) is bounded on \(A\).

Prove that if \(a>0, a \neq 1\), then \(a^{\log _{a} x}=x\) for all \(x \in(0, \infty)\) and \(\log _{a}\left(a^{y}\right)=y\) for all \(y \in \mathbb{R}\). Therefore the function \(x \mapsto \log _{a} x\) on \((0, \infty)\) to \(\mathbb{R}\) is inverse to the function \(y \mapsto a^{y}\) on \(\mathbb{R}\).

If \(x \geq 0\) and \(n \in \mathbb{N}\), show that $$ \frac{1}{x+1}=1-x+x^{2}-x^{3}+\cdots+(-x)^{n-1}+\frac{(-x)^{n}}{1+x} $$ Use this to show that $$ \ln (x+1)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\cdots+(-1)^{n-1} \frac{x^{n}}{n}+\int_{0}^{x} \frac{(-t)^{n}}{1+t} d t $$

Let \(f_{n}(x):=x / n\) for \(x \in[0, \infty), n \in \mathbb{N}\). Show that \(\left(f_{n}\right)\) is a decreasing sequence of continuous functions that converges to a continuous limit function, but the convergence is not uniform on \([0 . \infty)\)

If \(f: \mathbb{R} \rightarrow \mathbb{R}\) is such that \(f^{\prime \prime}(x)=f(x)\) for all \(x \in \mathbb{R}\), show that there exist real numbers \(\alpha, \beta\) such that \(f(x)=\alpha c(x)+\beta s(x)\) for all \(x \in \mathbb{R} .\) Apply this to the functions \(f_{1}(x):=e^{x}\) and \(f_{2}(x):=e^{-x}\) for \(x \in \mathbb{R} .\) Show that \(c(x)=\frac{1}{2}\left(e^{x}+e^{-x}\right)\) and \(s(x)=\frac{1}{2}\left(e^{x}-e^{-x}\right)\) for \(x \in \mathbb{R}\).
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