Problem 8 - Thomas Calculus

Question

Solve the equations. \(8^{\log _{8}(3)}-e^{\ln 5}=x^{2}-7^{\log _{7}(3 x)}\)

Step-by-Step Solution

Answer

The solutions are \(x = 1\) and \(x = 2\).

Step 1: Simplify both Logarithmic Expressions

Start with the expression on the left-hand side: - For the term \(8^{\log_{8}(3)}\), recognize that it's just 3, as raising a base to its logarithm returns the argument, thus \(8^{\log_{8}(3)} = 3\). - For the term \(e^{\ln(5)}\), similarly, it simplifies to 5, because \(e\) raised to its natural logarithm returns the argument. So, \(e^{\ln(5)} = 5\). Thus, the left-hand side simplifies to \(3 - 5 = -2\).

Step 2: Simplify the Second Logarithmic Expression

Now, let's address the right-hand side expression, \(x^{2} - 7^{\log_{7}(3x)}\).- For the term \(7^{\log_{7}(3x)}\), it simplifies to \(3x\) because the expression is in the form \(a^{\log_{a}(b)} = b\). So, \(7^{\log_{7}(3x)} = 3x\).

Step 3: Set Simplified Left and Right Expressions Equal

Now our equation reads:\[-2 = x^{2} - 3x\]Simplify by moving all terms to one side to form a quadratic equation:\[x^{2} - 3x + 2 = 0\]

Step 4: Solve the Quadratic Equation

We have the quadratic equation \(x^{2} - 3x + 2 = 0\). Factor it into two binomials:- Try finding numbers that multiply to 2 and add to -3. The numbers are -1 and -2.- Therefore, \(x^{2} - 3x + 2 = (x-1)(x-2) = 0\).Set each factor equal to zero:1. \(x - 1 = 0 \Rightarrow x = 1\)2. \(x - 2 = 0 \Rightarrow x = 2\)

Step 5: Verify Solutions

Test both solutions in the original equation:- For \(x = 1\), calculate both sides: - LHS: \(8^{\log_{8}(3)} - e^{\ln(5)} = 3 - 5 = -2\) - RHS: \(1^2 - 7^{\log_{7}(3 \times 1)} = 1 - 3 = -2\) - Both sides are equal, so \(x = 1\) is a solution.- For \(x = 2\), calculate both sides: - LHS: \(8^{\log_{8}(3)} - e^{\ln(5)} = 3 - 5 = -2\) - RHS: \(2^2 - 7^{\log_{7}(3 \times 2)} = 4 - 6 = -2\) - Both sides are equal, meaning \(x = 2\) is also a solution.

Key Concepts

Logarithmic Expressions

Logarithmic expressions involve logarithms, which are the inverses of exponential functions. Understanding logs can help simplify many mathematical equations. A logarithm answers the question: "To what power must the base be raised, to produce a given number?"

• For example, in the term \(8^{\log_{8}(3)}\), the base 8 is raised to the logarithm with the same base. This means the result is simply the argument 3, because \(a^{\log_a(b)} = b\).
• Similarly, with \(e^{\ln(5)}\), \(e\) and \(\ln\) are also inverse functions. The natural log \(\ln\) has a base \(e\), and thus \(e^{\ln(5)} = 5\).

These properties are vital for simplifying complex expressions and solving equations involving logarithms, as seen in this exercise.

Quadratic Equations

A quadratic equation is a type of polynomial equation of degree 2, usually in the form \(ax^{2} + bx + c = 0\). It can often be solved by factoring, using the quadratic formula, or completing the square.

• In our exercise, the quadratic equation formed was \(x^{2} - 3x + 2 = 0\).
• To solve a quadratic by factoring, we look for two numbers that multiply to the constant term \(c\) (here 2) and add up to the linear coefficient \(b\) (here -3).
• For \(x^{2} - 3x + 2\), these numbers are -1 and -2, leading to the factors \((x-1)(x-2) = 0\).

Setting each factor to zero gave us the solutions \(x = 1\) and \(x = 2\). Quadratic equations often have two solutions due to the nature of a parabola's shape.

Factoring Polynomials

Factoring polynomials involves expressing a polynomial as a product of its factors. This technique is frequently used to simplify expressions and solve equations.

• In this exercise, we factored the quadratic equation \(x^{2} - 3x + 2\) to find its roots.
• The polynomial is expanded into factors \((x-1)\) and \((x-2)\), which multiply to give the original expression \(x^{2} - 3x + 2\).
• Factoring is useful because it allows us to easily find the values of \(x\) that make the polynomial equal zero. These values, \(x = 1\) and \(x = 2\), are known as the roots or solutions.

Factoring is a powerful tool for simplifying and solving polynomial expressions, providing insight into the behavior of algebraic functions.