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Solve each equation. $$3(\ln x)^{2}-\ln \left(x^{2}\right)-8=0$$

Short Answer

Expert verified
The solutions are \(x = e^2\) and \(x = e^{-\frac{4}{3}}\).

Step by step solution

01

Simplify the equation

Start with the equation: \(3(\ln x)^2 - \ln(x^2) - 8 = 0\). Recognize that \(\ln(x^2) = 2\ln x\). Substitute this into the equation: \(3(\ln x)^2 - 2\ln x - 8 = 0\).
02

Use substitution to simplify further

Let \(y = \ln x\). This changes the equation to \(3y^2 - 2y - 8 = 0\).
03

Solve the quadratic equation

The quadratic equation \(3y^2 - 2y - 8 = 0\) can be solved using the quadratic formula: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -2\), and \(c = -8\).
04

Calculate the discriminant

Calculate the discriminant \(b^2 - 4ac = (-2)^2 - 4(3)(-8) = 4 + 96 = 100\).
05

Apply the quadratic formula

Using the quadratic formula, we have \(y = \frac{-(-2) \pm \sqrt{100}}{2 \times 3} = \frac{2 \pm 10}{6}\).
06

Find the roots of the quadratic equation

Calculate the roots: \(y_1 = \frac{2 + 10}{6} = 2\) and \(y_2 = \frac{2 - 10}{6} = -\frac{4}{3}\).
07

Solve for \(x\) in terms of \(\ln x\)

Since \(y = \ln x\), substitute back to solve for \(x\). For \(y_1 = 2\), \(\ln x = 2\), so \(x = e^2\). For \(y_2 = -\frac{4}{3}\), \(\ln x = -\frac{4}{3}\), so \(x = e^{-\frac{4}{3}}\).
08

Verify the solutions

Substitute \(x = e^2\) and \(x = e^{-\frac{4}{3}}\) back into the original equation to ensure they satisfy it. Both should satisfy \(3(\ln x)^2 - \ln(x^2) - 8 = 0\).

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Key Concepts

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

Exponential Functions
Exponential functions are fundamental in mathematics, especially when solving equations involving logarithms and powers. These functions take the shape of \( y = a^x \), where \( a \) is a constant and \( x \) is the exponent. In nature and science, exponential functions often model growth processes, like populations or bacteria, and decay processes, like radioactive materials.
  • An important feature of these functions is that they grow extremely fast, depending on the size of the exponent.
  • For instance, a base of \( e \) (approximated as 2.71828), known as Euler's number, is commonly used in calculus and natural logarithms.

Why Use Exponential Functions?

Exponential functions are versatile and show up in real-life scenarios, from calculating compound interest in finance to predicting population growth. If you ever encounter logarithmic equations, understanding exponential functions is a must, as logarithms are essentially the reverse operations.
So when you deal with logarithmic equations, like those involving \( \ln(x) \), converting them to exponential form can greatly simplify your work.
Quadratic Formula
The quadratic formula is an essential tool for solving quadratic equations, which are second-degree polynomials of the form \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are known coefficients, with \( x \) being the variable. The quadratic formula is expressed as: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Steps to Use the Quadratic Formula

  • Identify the coefficients \( a \), \( b \), and \( c \) from the equation.
  • Calculate the discriminant \( b^2 - 4ac \). The discriminant can tell us the nature of the roots:
    • If it is positive, there are two real and distinct roots.
    • If it is zero, there is one real repeated root.
    • If it is negative, the roots are complex or imaginary.
Plug in the values into the quadratic formula to find \( x \).
Understanding the quadratic formula is crucial for various algebraic computations, particularly in solving complex equations involving squared terms. In the exercise, substituting \( y = \ln x \) transformed the logarithmic problem into a quadratic equation, allowing us to apply the quadratic formula effectively.
Natural Logarithms
Natural logarithms are used extensively in mathematics and are the inverse of exponential functions with base \( e \). Notation for a natural logarithm is \( \ln(x) \), and it answers the question, "To what power must \( e \)be raised to obtain \( x \)?"

Importance of Natural Logarithms

  • They are particularly useful in calculus for integration and differentiation.
  • Many scientific processes, particularly those involving growth and decay, are naturally modeled on base \( e \).Thus, \( \ln \) functions are often found in formulas for these processes.
In the solved problem, recognizing that \( \ln(x^2) = 2\ln x \) played a vital role. The simplification step relied on this propertyof natural logarithms to convert the problem into an approachable quadratic equation.
By grasping natural logarithms, you can simplify complex logarithmic expressions, making them much easier to manage.This simplification is particularly beneficial when tackling mathematical scenarios requiring detailed analysis.

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Most popular questions from this chapter

The half-life of plutonium-241 is 13 years. (a) How much of an initial 2 -g sample remains after 5 years? (b) Find the time required for \(90 \%\) of the 2 -g sample to decay. Hint: If \(90 \%\) has decayed, then \(10 \%\) remains.

(a) Use a graphing utility to estimate the root(s) of the equation to the nearest one-tenth (as in Example 6). (b) Solve the given equation algebraically by first rewriting it in logarithmic form. Give two forms for each answer: an exact expression and a calculator approximation rounded to three decimal places. Check to see that each result is consistent with the graphical estimate obtained in part (a). $$10^{x^{2}}=40$$

Solve each equation and solve for \(x\) in terms of the other letters. $$T=T_{1}+\left(T_{0}-T_{1}\right) e^{-k x}$$

Solve the inequality \(\log _{10}\left(x^{2}-6 x-6\right)>0.\)

(a) Let \(\mathcal{N}(t)=\mathcal{N}_{0} e^{k t} .\) Show that \([\mathcal{N}(t+1)-\mathcal{N}(t)] / \mathcal{N}(t)=e^{k}-1 .\) (This is actually done in detail in the text. So, ideally, you should look back only if you get stuck or want to check your answer.) (b) Assume as given the following approximation, which was introduced in Exercise 26 of Section 5.2 \(e^{x} \approx x+1 \quad\) provided \(x\) is close to zero Use this approximation to explain why \(e^{k}-1 \approx k\) provided that \(k\) is close to zero. Remark: Combining this result with that in part (a), we conclude that the relative growth rate for the function \(\mathcal{N}(t)=\mathcal{N}_{0} e^{k t}\) is approximately equal to the growth constant \(k .\) As explained in the text, this is one of the reasons why in applications we've not distinguished between the relative growth rate and the decay constant \(k\)

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