Problem 22 - Technical Mathematics with Calculus

Question

Solve for \(x\). Give any approximate results to three significant digits. Check your answers. $$\ln x-2 \ln x=\ln 64$$

Step-by-Step Solution

Answer

The solution to the equation is \(x = 0.0156\), or to three significant digits, \(x = 0.0156\).

Step 1: Combine logarithmic terms

Combine the logarithmic terms on the left side of the equation using the law of logarithms that states \(\ln a - \ln b = \ln\frac{a}{b}\). In this case, substitute \(a\) with \(x\) and \(b\) with \(x^2\), due to the property \(2 \ln x = \ln x^2\). The equation becomes \(\ln\frac{x}{x^2}=\ln 64\).

Step 2: Simplify the logarithmic expression

Simplify the numerator and the denominator of the fraction inside the logarithm to get \(\ln\frac{1}{x}=\ln 64\).

Step 3: Apply the property of logarithmic equality

Since the base of the natural logarithm is \(e\), the property \(\ln a = \ln b \implies a = b\) applies. Set \(\frac{1}{x}\) equal to 64: \(\frac{1}{x} = 64\).

Step 4: Solve for x

To solve for \(x\), take the reciprocal of both sides of the equation: \(x = \frac{1}{64}\).

Step 5: Check the solution

Substitute \(x\) back into the original equation: \(\ln(\frac{1}{64}) - 2\ln(\frac{1}{64}) = \ln(64)\). Simplify the left side using logarithm properties \((\ln a - n\ln a = -(n-1)\ln a)\), and verify the right side is equivalent. Since \(\ln(64)\) is equal to \(\ln(64)\), the solution is correct.

Key Concepts

Natural Logarithm Properties

Understanding the properties of natural logarithms is essential to solve logarithmic equations effectively. The natural logarithm, denoted as \(\ln\), is a logarithm to the base \(e\), where \(e\) is an irrational constant approximately equal to 2.71828. Essential properties of natural logarithms include:

• \(\ln(1) = 0\) because \(e^0 = 1\).
• \(\ln(e) = 1\) since \(e^1 = e\).
• The logarithm of a product is the sum of the logarithms: \(\ln(ab) = \ln(a) + \ln(b)\).
• The logarithm of a quotient is the difference of the logarithms: \(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\).
• For a real number \(c\), the logarithm of a power is the product of \(c\) and the logarithm: \(\ln(a^c) = c\cdot\ln(a)\).

When solving logarithmic equations, these properties allow you to manipulate and simplify expressions to isolate the variable of interest.

Logarithmic Equality

Logarithmic equality is a principle that greatly simplifies the process of solving logarithmic equations. The concept behind it is straightforward: if two logarithms with the same base are equal, their arguments must also be equal. Mathematically, if you have \(\ln(a) = \ln(b)\), then it must follow that \(a = b\).

This property is derived from the definition of logarithms as inverses of exponential functions. In the context of natural logarithms, it means that if you have an equation \(\ln(x) = \ln(y)\), you can conclude that \(x = y\) because both \(x\) and \(y\) would be the result of raising \(e\) to the same power. Consequently, when faced with an equation such as \(\ln\left(\frac{x}{x^2}\right) = \ln(64)\), you can directly compare and solve for \(\frac{1}{x} = 64\) without having to deal with the logarithm anymore.

Simplifying Logarithmic Expressions

Simplification of logarithmic expressions is a valuable skill that helps in solving logarithmic equations. It involves applying logarithm properties to condense or expand logarithmic terms to a form that is easier to solve.

For instance, when you have multiple logarithmic terms such as \(\ln(x) - 2\ln(x)\), you can simplify using the property that equates multiplication inside the logarithm to addition or subtraction outside the logarithm. Converting the expression \(2\ln(x)\) to \(\ln(x^2)\) and then combining the terms to \(\ln\left(\frac{x}{x^2}\right)\) is a simplification that reduces the equation to a single logarithm, making it easier to equate and solve the argument of the logarithm.

When you further simplify \(\ln\left(\frac{x}{x^2}\right)\) to \(\ln\left(\frac{1}{x}\right)\), you demonstrate simplification by both combining like terms and reducing the fraction inside the logarithm. It's crucial to keep these properties in mind to streamline the process of solving logarithmic equations.