Problem 46 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 45-48: Solve the system for \(a\) and \(b\). (Hint: Treat terms such as \(e^{3 x}, \cos x\), and \(\sin x\) as "constant coefficients.") $$ \left\\{\begin{array}{r} a e^{-x}+b e^{4 x}=0 \\ -a e^{-x}+b\left(4 e^{4 x}\right)=2 \end{array}\right. $$

Step-by-Step Solution

Answer

\(a = -\frac{2 e^x}{5}\), \(b = \frac{2}{5 e^{4x}}\).

Step 1: Express One Equation in Terms of 'a'

Starting with the first equation of the system:\[ a e^{-x} + b e^{4x} = 0 \]Solve for \(a\):\[ a e^{-x} = -b e^{4x} \]\[ a = -b e^{4x} \cdot e^{x} \]\[ a = -b e^{5x} \]

Step 2: Substitute 'a' in the Second Equation

Substitute the expression for \(a\) from Step 1 into the second equation:\[ -a e^{-x} + b(4 e^{4x}) = 2 \]Substitute \(a = -b e^{5x} \):\[ -(-b e^{5x}) e^{-x} + 4b e^{4x} = 2 \]Simplify the left side:\[ b e^{4x} + 4b e^{4x} = 2 \]\[ 5b e^{4x} = 2 \]

Step 3: Solve for 'b'

From \(5b e^{4x} = 2\), solve for \(b\):\[ b = \frac{2}{5 e^{4x}} \]

Step 4: Solve for 'a' Using b's Value

Now that we have \(b\), substitute back into the expression for \(a\) from Step 1:\[ a = -b e^{5x} \]Substitute \(b = \frac{2}{5 e^{4x}} \):\[ a = -\frac{2}{5 e^{4x}} \cdot e^{5x} \]\[ a = -\frac{2 e^{5x}}{5 e^{4x}} \]\[ a = -\frac{2 e^x}{5} \]Thus, the values are \(a = -\frac{2 e^x}{5}\) and \(b = \frac{2}{5 e^{4x}}\).

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent, typically written as \(f(x) = a \cdot b^x\), where \(a\) is a constant coefficient, \(b\) is the base, and \(x\) is the exponent.
In our given system, we encounter terms like \(e^x\) and \(e^{4x}\), where \(e\) is the base of the natural logarithm, approximately equal to 2.718.
Exponential functions often appear in contexts such as compound interest, growth/decay modeling, and differential equations.
### Key Properties of Exponential Functions - **Growth/Decay:** Exponential functions grow or decay rapidly based on the exponent. For positive exponents, the function grows, while for negative ones, it decays.- **Derivative:** The derivative of \(e^x\) is \(e^x\) itself, making it unique in calculus and key in solving differential equations.- **Integration:** The integral of \(e^x\) also returns \(e^x\), plus a constant of integration, which plays a role in solving antiderivatives.
Understanding exponential functions is crucial for solving equations like our exercise's system, as it involves manipulating these terms systematically.

Solving Systems

Solving systems of equations involves finding the values of variables that satisfy multiple equations simultaneously.
This method is crucial because many real-world problems require solving interconnected relationships, represented by such systems.
### Two Main Methods for Solving Systems - **Substitution:** Solve one equation for a variable and substitute this into the other equation, reducing the system to a single equation.- **Elimination:** Add or subtract equations to eliminate one of the variables, simplifying the system to solve for the remaining variables.
In our exercise, substitution was used. First, one equation was solved for \(a\), which was then used to replace \(a\) in the second equation. This reduced the system to one solvable equation in terms of \(b\), which was then back-substituted to find \(a\).
Understanding these methods is essential for tackling diverse problems, giving us the tools to decipher complex relationships between variables.

Algebraic Manipulation

Algebraic manipulation refers to the use of algebraic techniques to rearrange and simplify equations to solve for unknowns.
It involves operations like combining like terms, factoring, expanding expressions, and isolating variables.
### Common Techniques in Algebraic Manipulation - **Factoring and Expanding:** Helps in simplifying polynomials and expressions, such as rewriting \(ax^2 + bx + c\) in factored form.- **Combining Like Terms:** Allows the consolidation of terms with the same variable powers, which simplifies equations.- **Isolating Variables:** Used to express a variable in terms of others, which is crucial when solving algebraic equations.
In the provided system, algebraic manipulation played a role by rewriting expressions to isolate \(a\) and \(b\). For instance, by expressing \(a\) in terms of \(b\), then substituting in subsequent steps, we could systematically arrive at solutions.
Mastery of these manipulation techniques is vital for simplifying and solving complex problems efficiently.