/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 109 Describe the change-of-base prop... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 109

Describe the change-of-base property and give an example.

Short Answer

Expert verified
The change-of-base property for logarithms states that \(\log_b x = \frac{\log_a x}{\log_a b}\) for any positive numbers \(a\), \(b\), \((b鈮�1)\) and \(x\). This property helps to calculate logarithms with bases that are not easily computable. An example of this: \(\log_{2} 64 = \frac{\log_ {10} 64}{\log_ {10} 2} 鈮� 6\).

Step by step solution

01

Understand the change-of-base property

The change-of-base property of logarithms states that for any positive number \(a\), \(b\), \((b鈮�1)\) and \(x\), the logarithm base \(b\) of \(x\) can be computed with the help of logarithms with another base \(a\). This is especially helpful when dealing with bases that are not on a calculator. It is generally written as: \(\log_b x = \frac{\log_a x}{\log_a b}\). Note that the base \(a\) in the new expression can be any positive value except 1.
02

Provide the derivation of the formula

Suppose \(y = b^x\), then take the logarithm of both sides using the base \(a\), we will get \(\log_a y = x \cdot log_a b\). if \(y=x\), we will get \(\log_a x = x \cdot log_a b\) hence, \(x = \frac{\log_a x}{\log_a b}\). This is how we can arrive at the change-of-base formula.
03

Example using change-of-base property

For example, consider calculating \(\log_{2} 64\). Using the change-of-base property, this can transform into \(\frac{\log_ {10} 64}{\log_ {10} 2}\), which are values we can easily calculate with a calculator. So, \(\log_{2} 64 = \frac{\log_ {10} 64}{\log_ {10} 2} = \frac{1.806}{0.301} 鈮� 6\).

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Most popular questions from this chapter

The loudness level of a sound, \(D,\) in decibels, is given by the formula $$D=10 \log \left(10^{12} I\right)$$ where \(I\) is the intensity of the sound, in watts per meter \(^{2}\). Decibel levels range from \(0,\) a barely audible sound, to \(160,\) a sound resulting in a ruptured eardrum. (Any exposure to sounds of 130 decibels or higher puts a person at immediate risk for hearing damage.) Use the formula to solve. What is the decibel level of a normal conversation, \(3.2 \times 10^{-6}\) watt per meter \(^{2} ?\)

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Write each equation in its equivalent exponential form. Then solve for \(x .\) $$ \log _{4} x=-3 $$
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