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Chapter 3: Problem 32

Write each exponential equation in its equivalent logarithmic form. $$e^{-x}=4$$

Short Answer

Expert verified
The logarithmic form is \( \ln(4) = -x \).

Step by step solution

01

Identify the Exponential Equation

We're given the exponential equation: \( e^{-x} = 4 \). This equation involves a base \( e \), which is the natural exponential base.
02

Understand Exponential to Logarithmic Conversion

To write an exponential equation \( a^b = c \) in logarithmic form, we apply the conversion formula: \( \log_a(c) = b \). Here, \( a \) is the base, \( c \) is the result, and \( b \) is the exponent.
03

Apply the Conversion to the Given Equation

For the equation \( e^{-x} = 4 \), the base \( a \) is \( e \), the result \( c \) is 4, and the exponent \( b \) is \(-x\). Thus, the equivalent logarithmic form is: \( \ln(4) = -x \), where \( \ln \) denotes the natural logarithm function.

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Key Concepts

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

Logarithmic Form
Logarithmic form is a different way of expressing an exponential equation. When you have an exponential equation like \( a^b = c \), you can convert it into a logarithm. This alternative form gives you the expression \( \log_a(c) = b \).

In this setup:
  • \(a\) is the base.
  • \(c\) is known as the result or the power.
  • \(b\) is the exponent.
Using the logarithmic form is helpful because it allows you to solve for the exponent. It's frequently used in mathematics because it simplifies complex multiplicative relationships into more manageable additive ones.
Natural Logarithms
Natural logarithms utilize the number \(e\) as their base. The number \(e\) is an important mathematical constant, approximately equal to 2.71828. When you see \(\ln(x)\), it's asking for the power to which \(e\) must be raised to result in \(x\).

Here are some helpful points about natural logarithms:
  • \(\ln(e) = 1\), because \(e^1 = e\).
  • \(\ln(1) = 0\), since any number to the power of zero is 1.
  • Natural logarithms are integral to calculus, owing to their nice derivative and integral properties.
Knowing how to work with \(\ln\) is essential for solving equations where the base \(e\) is involved.
Exponential to Logarithmic Conversion
Converting an exponential equation to its logarithmic form is a straightforward process that plays a key role in solving various mathematical problems. When you have an equation expressed exponentially like \(e^{-x} = 4\), you apply the rule of conversion. The general rule is \(a^b = c\) can be rewritten in logarithmic form as \(\log_a(c) = b\).

For the equation \(e^{-x} = 4\):
  • The base \(a\) is the natural number \(e\).
  • The result \(c\) is 4.
  • The exponent \(b\) is \(-x\).
So, the conversion gives you \(\ln(4) = -x\). Understanding this process helps in problem-solving as it allows you to isolate the variable \(x\) and find its value more easily.

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Most popular questions from this chapter

Refer to the following: In calculus, we find the derivative, \(f^{\prime}(x),\) of a function \(f(x)\) by allowing \(h\) to approach 0 in the difference quotient \(\frac{f(x+h)-f(x)}{h}\) of functions involving exponential functions. Use the fact that \(\frac{e^{h}-1}{h}=1\) when \(h\) is close to zero to find the derivative of \(f(x)=e^{x}+x\).

Determine whether each statement is true or false. $$e^{\log x}=x$$

Recall that the derivative of \(f\) can be found by letting \(h \rightarrow 0\) in the difference quotient \(\frac{f(x+h)-f(x)}{h} .\) In calculus we prove that \(\frac{e^{h}-1}{h}=1,\) when \(h\) approaches \(0 ;\) that is, for really small values of \(h, \frac{e^{h}-1}{h}\) gets very close to 1. Use this information to find the derivative of \(f(x)=e^{x}\).

Wing Shan just graduated from dental school owing \(\$ 80,000\) in student loans. The annual interest is \(6 \% .\) Her time \(t\) to pay off the loan is given by $$t=-\frac{\ln \left[1-\frac{80,000(0.06)}{n R}\right]}{n \ln \left(1+\frac{0.06}{n}\right)}$$ where \(n\) is the number of payment periods per year and \(R\) is the periodic payment. a. Use a graphing utility to graph $$t_{1}=-\frac{\ln \left[1-\frac{80,000(0.06)}{12 x}\right]}{12 \ln \left(1+\frac{0.06}{12}\right)} \text { as } Y_{1} \text { and }$$ $$t_{2}=-\frac{\ln \left[1-\frac{80,000(0.06)}{26 x}\right]}{26 \ln \left(1+\frac{0.06}{26}\right)} \text { as } Y_{2}$$ Explain the difference in the two graphs. b. Use the \([\text { TRACE }]\) key to estimate the number of years that it will take Wing Shan to pay off her student loan if she can afford a monthly payment of \(\$ 800\) c. If she can make a biweekly payment of \(\$ 400\), estimate the number of years that it will take her to pay off the loan. d. If she adds \(\$ 200\) more to her monthly or \(\$ 100\) more to her biweekly payment, estimate the number of years that it will take her to pay off the loan.

Determine whether each statement is true or false. The division of two logarithms with the same base is equal to the logarithm of the subtraction.
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