Problem 10 - Thomas Calculus

Question

Solve the equations. \(\ln e+4^{-2 \log _{4}(x)}=\frac{1}{x} \log _{10}(100)\)

Step-by-Step Solution

Answer

The solution to the equation is \( x = 1 \).

Step 1: Simplify Components

Start by considering each part of the equation independently. Simplify any constants or commonly known values. - The term \( \ln e = 1 \), because the natural logarithm of \( e \) is always 1.- Recognize that \( \log_{10}(100) = 2 \), since 100 is 10 squared.

Step 2: Simplify the Equation

Substitute the simplified terms back into the equation:\[ 1 + 4^{-2 \log_{4}(x)} = \frac{2}{x} \]Our task now is to simplify the exponential term \( 4^{-2 \log_{4}(x)} \).

Step 3: Simplify the Power of 4 Term

Utilize the property of exponents: \( a^{\log_{a}(b)} = b \). Here, rewrite the power term:\[ 4^{-2 \log_{4}(x)} = (4^{-2})^{\log_{4}(x)} = \left(\frac{1}{16}\right)^{\log_{4}(x)} \]Let \( y = 4^{\log_{4}(x)} = x \), implying \( y^{-2} = x^{-2} \), thus simplifying \( 4^{-2 \log_{4}(x)} \) to \( x^{-2} \).

Step 4: Substitute Simplified Power Term

Substitute \( 4^{-2 \log_{4}(x)} = x^{-2} \) into the equation:\[ 1 + x^{-2} = \frac{2}{x} \]

Step 5: Solve the Resulting Equation

Clear the fractions by multiplying through by \( x^2 \) to eliminate the denominators:\[ x^2 + 1 = 2x \]Rearrange this into a quadratic equation:\[ x^2 - 2x + 1 = 0 \]

Step 6: Solve the Quadratic Equation

Recognize that the quadratic equation \( x^2 - 2x + 1 = 0 \) can be factored:\[ (x - 1)^2 = 0 \]So, \( x - 1 = 0 \) leading to \( x = 1 \).

Step 7: Check the Solution

Verify the solution by substituting \( x = 1 \) back into the original equation:- The left side becomes: \( \ln e + 4^{-2 \log_{4}(1)} = 1 + 4^0 = 2 \).- The right side becomes: \( \frac{1}{1} \cdot 2 = 2 \).Both sides are equal, confirming \( x = 1 \) is the correct solution.

Key Concepts

Natural Logarithm

The natural logarithm is a fundamental concept in mathematics, widely used for solving logarithmic equations. Denoted as \(\ln\), this logarithm has a base \(e\), where \(e\) is an irrational constant approximately equal to 2.71828. The natural logarithm of a number, for instance, \(\ln(x)\), represents the power to which you must raise \(e\) to get \(x\).
In simple terms, when you see \(\ln(e)\), it simplifies directly to 1. This is because \(e\) raised to the power of 1 is \(e\). This property simplifies calculations and is crucial in solving equations that involve \(e\), just like in the presented exercise.
If you encounter a problem with a natural logarithm, identifying these simplifications first will guide you more effectively to the solution.

Logarithm Properties

Logarithms have several properties that simplify mathematical expressions, especially when dealing with equations. Here are the key properties:

• Product Property: \(\log_b(mn) = \log_b(m) + \log_b(n)\)
• Quotient Property: \(\log_b\left(\frac{m}{n}\right) = \log_b(m) - \log_b(n)\)
• Power Property: \(\log_b(m^n) = n \cdot \log_b(m)\)
• Change of Base Formula: \(\log_b(m) = \frac{\log_k(m)}{\log_k(b)}\)

These properties ease the simplification of complex logarithmic expressions.
In the exercise above, understanding that \(\log_{10}(100) = 2\) is a direct application of the power property. It transforms the equation to manageable terms, allowing easier progress towards finding the solution.

Exponentiation

Exponentiation is a mathematical operation involving two numbers, the base and the exponent. The base raised to the power of the exponent represents repeated multiplication of the base. For example, \(4^{-2}\) is equivalent to \(\frac{1}{16}\), as 4 raised to the negative power gives the reciprocal.
Exponentiation also has specific properties, including:

• \(a^0 = 1\) for any non-zero \(a\).
• \(a^{-m} = \frac{1}{a^m}\).
• \(a^{m+n} = a^m \cdot a^n\).

The crucial part of the original solution involves recognizing that exponentiation combined with logarithms can often simplify to straightforward expressions. The property \(a^{\log_a(b)} = b\) is used to transform the given equation parts, making them more solvable. Understanding how exponentials invert logarithms allows for elegant simplification.

Quadratic Equations

Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable. These equations can be solved using various methods:

• Factoring: Expressing the quadratic in a product form, such as \((x-h)(x-k) = 0\), and solving for \(x\).
• Using the Quadratic Formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• Completing the Square: Transforming the equation into a perfect square trinomial.

In the provided solution, the quadratic \(x^2 - 2x + 1 = 0\) was factored into \((x-1)^2 = 0\), allowing a simple solution by recognizing it as a perfect square trinomial. This approach is straightforward when the quadratic can be easily transformed by recognizing patterns or through factorization.