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Chapter 9: Problem 58

Solve each equation. Round to the nearest ten-thousandth. \(\ln 4 x+\ln x=9\)

Short Answer

Expert verified
\(x \approx 45.0119\)

Step by step solution

01

Combine Logarithms

Use the logarithm property that states \(\ln a + \ln b = \ln(ab)\) to combine the logarithms. This gives us: \[\ln (4x \cdot x) = \ln (4x^2) = 9\]
02

Exponentiate Both Sides

Remove the logarithm by exponentiating both sides of the equation. Since \(e^{\ln x} = x\), we have:\[\ln (4x^2) = 9 \rightarrow 4x^2 = e^9\]
03

Solve for \(x^2\)

Isolate \(x^2\) by dividing both sides by 4:\[x^2 = \frac{e^9}{4}\]
04

Calculate \(e^9\)

Use a calculator to find \(e^9\) (approximately \(8103.0839\)), and then divide by 4 to find \(x^2\):\[x^2 = \frac{8103.0839}{4} = 2025.770975\]
05

Take the Square Root

Find \(x\) by taking the square root of \(2025.770975\):\[x \approx \sqrt{2025.770975} \approx 45.0119\]
06

Round the Answer

Round \(x\) to the nearest ten-thousandth as required by the problem. Therefore, \[x \approx 45.0119\]

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Key Concepts

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

Understanding Combining Logarithms
Combining logarithms is a handy tool to simplify expressions like \( \ln 4x + \ln x \). The property used here is: **\( \ln a + \ln b = \ln(ab)\)**. This means the sum of two logarithms can be represented as the logarithm of a single product.

This property helps us shift from multiple terms to a single term, making it much easier to solve equations. In the given problem, when we combine \( \ln 4x + \ln x \), it becomes \( \ln (4x^2) \). Always remember, the base of the logarithms must be the same for this technique to work.
Exploring Exponentiation
Exponentiation is a process that helps in removing logarithms from an equation. For natural logarithms, you use the fact that \( e^{\ln x} = x \).

In our case, if we have \( \ln(4x^2) = 9 \), we exponentiate both sides: **\( e^{\ln(4x^2)} = e^9 \)**. The left simplifies to \( 4x^2 \), removing the logarithm entirely, so you have \( 4x^2 = e^9 \).

Exponentiation essentially uses the power of the exponential function (often base \( e \) in natural logarithms) to simplify equations for easier manipulation.
The Process of Solving Quadratic Equations
Once you have simplified to an equation like \( 4x^2 = e^9 \), solving for \( x \) involves solving a quadratic equation. Here, you first isolate \( x^2 \) by dividing by 4 to get: \[x^2 = \frac{e^9}{4}\]This is your typical quadratic form where you solve by taking the square root of both sides.

Generally, solving quadratic equations involves finding numbers that satisfy \( ax^2 + bx + c = 0 \). Although our problem doesn't have a linear term (like \( bx \)), the process of isolating \( x \) remains similar.
Achieving Numerical Approximation
In mathematics, solutions often require approximation, facilitating practical computation. Once the expression is simplified, such as \( x^2 = \frac{8103.0839}{4} \), the next step is numerical approximation.

Calculators or computer software can be used to find approximate values. In this case, calculate \( e^9 \) (which is approximately \( 8103.0839 \)), divide by 4 -> yielding \( 2025.770975 \), and take the square root. This approximates \( x \approx 45.0119 \).

Numerical approximations round results to a designated level of precision, here to the nearest ten-thousandth, which is an essential step in ensuring numeric clarity and correctness in real-world computations.

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Most popular questions from this chapter

Write an equivalent exponential or logarithmic equation. $$ e^{3}=y $$

Use \(\log _{7} 2 \approx 0.3562\) and \(\log _{7} 3 \approx 0.5646\) to approximate the value of each expression. $$ \log _{7} 27 $$

Solve each equation or inequality. Round to the nearest ten-thousandth. \(e^{x}>1.6\)

PREREQUISITE SKILL Solve each equation or inequality. Check your solutions. $$ \log _{10} 2^{x}=\log _{10} 32 $$

Solve each equation. Round to the nearest hundredth. \(10(1+0.25)^{x}=200\)
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