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Chapter 3: Problem 8

Write an equivalent exponential equation. $$-\log _{b} V=w$$

Short Answer

Expert verified
\(b^{-w} = V\)

Step by step solution

01

Understand Logarithmic and Exponential Forms

Recall that \(\text{if} \log_b(x) = y\), then \(b^y = x\). The logarithmic form and the exponential form are inverses of each other.
02

Rewrite the Given Logarithmic Equation

The given equation is \(-\log_b V = w\). To make it easier to convert to exponential form, isolate the logarithm: \minus(-\log_b V = w)\ implies \(\log_b V = -w\).
03

Convert to Exponential Form

Using the rule that \(\log_b(x) = y\) means \(b^y = x\), apply this to the rearranged equation \(\log_b V = -w\). This gives \(b^{-w} = V\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Logarithmic Form
Logarithmic form helps us understand the relationship between numbers and their powers. It's written as \(\text{\log}_b(x) = y\). In plain language, this means 'What exponent do we need on base b to get x.' Imagine it like finding the missing piece in a multiplication puzzle. Here, \(b\) is the base, \(x\) is the result, and \(y\) is the exponent.

For example, if \(\text{\log}_2(8) = 3\), it means that you need to raise 2 to the power of 3 to get 8.

The logarithmic form is handy because it allows us to work backwards from the result to find the original base and exponent. It鈥檚 the opposite of raising a number to a power.
Exponential Form
Exponential form is another way to describe the relationship between numbers. It鈥檚 written as \(b^y = x\). This states that if you take a base \(b\) and raise it to a power \(y\), you鈥檒l get \(x\).

For example, \(2^3 = 8\) means that 2 raised to the power of 3 equals 8.

Converting between exponential and logarithmic forms hinges on recognizing that they are just different perspectives of the same relationship. If you know \(b^y = x\), you can always reframe it as \(\text{\log}_b(x) = y\). This makes solving different kinds of problems much easier since you can utilize the format that suits the problem best.
Logarithmic Equations
Logarithmic equations include a logarithm of a variable. They follow the general form \(\text{\log}_b(x) = y\). To solve for the variable, you often convert the log form to exponential form. This helps to simplify and solve the equation.

Here鈥檚 the exercise example explained:
  • Start with \(-\text{\log}_b(V) = w\).
  • Isolate the logarithm: \(\text{\log}_b(V) = -w\).
  • Apply the conversion rule: \(\text{\log}_b(V) = -w\) means \(b^{-w} = V\).

This process shows how to switch from logarithmic to exponential form and solve for the original variable easily. By understanding these principles, you can tackle a variety of logarithmic and exponential problems more confidently.

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Most popular questions from this chapter

It is known that \(45 \%\) of all aluminum cans distributed will be recycled each year. A beverage company uses 250,000 lb of aluminum cans. After recycling, the amount of aluminum, in pounds, still in use after \(t\) years is given by $$N(t)=250,000(0.45)^{t}$$ (Source: The Container Recycling Institute.) a) Find \(N^{\prime}(t)\) b) Interpret the meaning of \(N^{\prime}(t)\)

A quantity \(Q_{1}\) grows exponentially with a doubling time of 1 yr. A quantity \(Q_{2}\) grows exponentially with a doubling time of 2 yr. If the initial amounts of \(Q_{1}\) and \(Q_{2}\) are the same, how long will it take for \(Q_{1}\) to be twice the size of \(Q_{2} ?\)

Differentiate. $$y=6^{x} \cdot \log _{7} x$$

Use the Chain Rule, implicit differentiation, and other techniques to differentiate each function given. $$y=2^{x^{4}}$$

Suppose that \(\$ 100\) is invested at \(7 \%,\) compounded continuously, for 1 yr. We know from Example 4 that the ending balance will be \(\$ 107.25 .\) This would also be the ending balance if \(\$ 100\) were invested at 7.25 \(\%,\) compounded once a year (simple interest). The rate of \(7.25 \%\) is called the effective annual yield. In general, if \(P_{0}\) is invested at interest rate \(k,\) compounded continuously, then the effective annual yield is that number i satisfying \(P_{0}(1+i)=P_{0} e^{k} .\) Then, \(1+i=e^{h},\) or Effective annual yield \(=i=e^{k}-1\) The effective annual yield on an investment compounded continuously is \(6.61 \% .\) At what rate was it invested?
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