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Chapter 1: Problem 24

Solve for \(x\) without using a calculating utility. Use the natural logarithm anywhere that logarithms are needed. $$3^{x}=2$$

Short Answer

Expert verified
\(x \approx 0.631\)

Step by step solution

01

Rewrite the Exponential Equation

Start with the given equation: \(3^x = 2\). To solve for \(x\), we'll need to involve logarithms because we are dealing with an exponential equation.
02

Apply Natural Logarithm

Apply the natural logarithm (\(\ln\)) to both sides of the equation to bring the exponent down: \(\ln(3^x) = \ln(2)\).
03

Utilize Logarithm Property

Use the logarithm power rule, \(\ln(a^b) = b \ln(a)\), to simplify the left-hand side: \(x \ln(3) = \ln(2)\).
04

Solve for x

Isolate \(x\) by dividing both sides of the equation by \(\ln(3)\): \(x = \frac{\ln(2)}{\ln(3)}\).

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Key Concepts

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

Natural Logarithms
The natural logarithm, denoted as \( \ln \), is a special kind of logarithm with a base of \( e \), where \( e \) is approximately 2.718. It is widely used in various mathematical calculations because it simplifies many expressions involving exponential growth or decay.

When you see \( \ln(x) \), it means "What power must we raise \( e \) to in order to get \( x \)?" The natural logarithm has a direct relationship with exponentials, making it a perfect tool for dealing with exponential equations.
  • Often used in continuous growth problems.
  • It allows us to easily manipulate and solve equations where the variable is an exponent.

Understanding \( \ln \) is crucial because it helps convert exponential equations into linear ones, simplifying the solving process.
Properties of Logarithms
Logarithms, in general, have several essential properties that simplify complex expressions or equations. One key property is the logarithm power rule. It states that \( \ln(a^b) = b \ln(a) \), which allows an exponent to "come down" and become a multiple of the logarithm.

This property is particularly useful when dealing with exponential expressions. For example, if you have \( 3^x = 2 \) and apply \( \ln \) to both sides, you leverage this rule: \( \ln(3^x) = x \ln(3) \).
  • Transforms products into sums: \( \ln(a \times b) = \ln(a) + \ln(b) \).
  • Turns divisions into differences: \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \).

Overall, these properties facilitate easier manipulation of logarithmic and exponential equations, allowing us to isolate variables and solve for them neatly.
Solving Logarithmic Equations
Logarithmic equations can present a challenge, but with a clear approach, they can be tackled effectively. Suppose you encounter an equation like \( 3^x = 2 \).

Here is how to break it down:
  • Start with the equation: When given an exponential equation, the variable is in the exponent, such as in \( 3^x = 2 \).
  • Apply logarithms: Use the natural logarithm on both sides, \( \ln(3^x) = \ln(2) \), to "bring down" the exponent for easier manipulation.
  • Use logarithmic properties: The power rule allows you to express it as \( x \ln(3) = \ln(2) \) where the variable \( x \) is now a multiplier.
  • Solve for the variable: Isolate \( x \) by dividing both sides by \( \ln(3) \), yielding \( x = \frac{\ln(2)}{\ln(3)} \).

Following this process simplifies solving such equations, ensuring a clear path to isolating the variable and finding its value without needing complex computational tools.

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Most popular questions from this chapter

(a) Find the smallest positive number \(N\) such that for each \(x\) in the interval \((N,+\infty),\) the value of the function \(f(x)=1 / x^{2}\) is within 0.1 unit of \(L=0\) (b) Find the smallest positive number \(N\) such that for each \(x\) in the interval \((N,+\infty),\) the value of \(f(x)=\) \(x /(x+1)\) is within 0.01 unit of \(L=1\) (c) Find the largest negative number \(N\) such that for each \(x\) in the interval \((-\infty, N),\) the value of the function \(f(x)=1 / x^{3}\) is within 0.001 unit of \(L=0\) (d) Find the largest negative number \(N\) such that for each \(x\) in the interval \((-\infty, N)\), the value of the function \(f(x)=x /(x+1)\) is within 0.01 unit of \(L=1\)

The population \(p\) of the United States (in millions) in year \(t\) can be modeled by the function $$p(t)=\frac{525}{1+1.1 e^{-0.02225(t-1990)}}$$ (a) Based on this model, what was the U.S. population in \(1990 ?\) (b) Plot \(p\) versus \(t\) for the 200 -year period from 1950 to 2150 (c) By evaluating an appropriate limit, show that the graph of \(p\) versus \(t\) has a horizontal asymptote \(p=c\) for an appropriate constant \(c\) (d) What is the significance of the constant \(c\) in part (c) for the population predicted by this model?

Use the Squeezing Theorem to show that $$\lim _{x \rightarrow 0} x \cos \frac{50 \pi}{x}=0$$ and illustrate the principle involved by using a graphing utility to graph the equations \(y=|x|, y=-|x|,\) and \(y=x \cos (50 \pi / x)\) on the same screen in the window \([-1,1] \times[-1,1]\).

(a) Make a conjecture about the general shape of the graph of \(y=\log (\log x),\) and sketch the graph of this equation and \(y=\log x\) in the same coordinate system. (b) Check your work in part (a) with a graphing utility.

(a) Show that $$\left|\left(3 x^{2}+2 x-20\right)-300\right|=|3 x+32| \cdot|x-10|$$ (b) Find an upper bound for \(|3 x+32|\) if \(x\) satisfies \(|x-10|<1\) (c) Fill in the blanks to complete a proof that $$\lim _{x \rightarrow 10}\left[3 x^{2}+2 x-20\right]=300$$ Suppose that \(\epsilon>0 .\) Set \(\delta=\min (1, \longrightarrow)\) and assume that \(0<|x-10|<\delta .\) Then $$\begin{aligned}\left|\left(3 x^{2}+2 x-20\right)-300\right| &=|3 x+32| \cdot|x-10| \\ &<\quad|x-10| =\epsilon\end{aligned}$$
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