Problem 55 - Algebra and Trigonometry

Question

Follow the hints and solve the systems. (a) \(\left\\{\begin{array}{cc}{\log x+\log y} & {=\frac{3}{2}} \\ {2 \log x-\log y} & {=0}\end{array} \quad[\text { Hint: Add the equations. }]\right.\) (b) \(\left\\{\begin{array}{ll}{2^{x}+2^{y}=10} & {\text { [Hint: Note that }} \\\ {4^{x}+4^{y}=68} & {4^{x}=2^{2 x}=\left(2^{x}\right)^{2} . ]}\end{array}\right.\) (c) \(\left\\{\begin{array}{rlrl}{x-y} & {=3} & {} & {\text { [Hint: Factor the left side }} \\ {x^{3}-y^{3}} & {=387} & {} & {\text { of the second equation. } ]}\end{array}\right.\) (d) \(\left\\{\begin{array}{ll}{x^{2}+x y=1} & {\text { [Hint: Add the equations }} \\ {x y+y^{2}=3} & {\text { and factor the result. } ]}\end{array}\right.\)

Step-by-Step Solution

Answer

(a) \(x^3 = 10\sqrt{10}, y = x^2\); (b) Solve for \(a, b\); (c) \(x^2 + xy + y^2 = 129\); (d) \((x+y)^2 = 4\).

Step 1: Simplify Equations

For equation (a), we have:\[ \log x + \log y = \frac{3}{2} \]and\[ 2 \log x - \log y = 0 \]We can rewrite these as:\[ \log xy = \frac{3}{2} \]and\[ \log x^2 = \log y \]

Step 2: Solve System Using Hints

For (a), add the equations:\[ \log xy + 2\log x - \log y = \frac{3}{2} + 0 \]which simplifies to:\[ \log(x^3) = \frac{3}{2} \]Then:\[ x^3 = 10^{\frac{3}{2}} = \sqrt{10^3} = \sqrt{1000} = 10\sqrt{10} \]And since \( \log x^2 = \log y \), we have:\[ y = x^2 \]

Step 3: Solve for Variables in Terms of Powers

For (b), let \( a = 2^x \) and \( b = 2^y \). Then we have:\[ a + b = 10 \]and\[ 4^x + 4^y = a^2 + b^2 = 68 \]

Step 4: Use Hints to Solve Cubic Relations

For (c), factor the second equation as:\[ (x-y)(x^2 + xy + y^2) = x^3 - y^3 = 387 \]Substitute \( x-y = 3 \):\[ 3(x^2 + xy + y^2) = 387 \]\[ x^2 + xy + y^2 = 129 \]

Step 5: Combine and Factor Quadratics

For (d), add the equations:\[ x^2 + xy + xy + y^2 = 1 + 3 \]Which simplifies to:\[ x^2 + 2xy + y^2 = 4 \]Factoring gives:\[ (x+y)^2 = 4 \]so \( x+y = 2 \) or \( x+y = -2 \).

Key Concepts

Logarithmic Equations

Logarithmic equations involve variables inside logarithm functions. They are equations where the unknown appears as the argument of a logarithm. To solve them, we often use logarithmic properties such as the product rule, power rule, and change of base.

• Product Rule: \(\log_a (b\cdot c) = \log_a b + \log_a c\)
• Power Rule: \(\log_a (b^c) = c \cdot \log_a b\)
• Change of Base Formula: \(\log_a b = \frac{\log_c b}{\log_c a}\)

In the given exercise (a), we could write the equations using the properties of logarithms. The goal was to simplify and solve for the variables by combining the logarithms into a single expression. By following the hints and algebraic manipulation, we use the property of logs to find relationships between the variables.

Exponential Equations

Exponential equations are ones where variables appear in the exponent. These equations often involve a constant base raised to a variable power. Two common forms are \(a^x = b\) and \(a^{f(x)} = b\). To solve them, we can use logarithms to bring the exponents down or express the base in a common form.

• Rewriting the Base: Sometimes these equations can be simplified by expressing the base as a power of each other, like changing \(4 = 2^2\) in the exercise.
• Using Logarithms: Applying logs on both sides of an equation can help solve for the exponent.

In exercise (b), we're introduced to compound expressions using exponential equations. We solved these by substituting a term (i.e., let \(a = 2^x\)) to easily solve the system and find the value of each variable.

Cubic Equations

Cubic equations are polynomial equations of the form \(ax^3 + bx^2 + cx + d = 0\). Solving them involves different strategies compared to linear or quadratic equations. In some cases, factorizations and substitutions simplify these equations.

• Factorization: Useful when you can rewrite the equation as a product of polynomials, as seen in the exercise when \(x-y = 3\) was substituted in.
• Roots of Equations: Identifying possible roots using special values or the Rational Root Theorem can help.

For exercise (c), recognizing that \(x-y\) was a shared component allowed us to break down the cubic difference into simpler, solvable steps. Factorizing helped us solve the complex part of the equations systematically.

Quadratic Equations

Quadratic equations are second-degree polynomials of the form \(ax^2 + bx + c = 0\). They are often solved by factoring, completing the square, or using the quadratic formula. When dealing with systems, these equations can be manipulated to reveal solutions for \(x\) and \(y\).

• Standard Form & Roots: Sometimes, equations are re-framed into \(\text{(quadratic)} = 0\), where we find roots using the discriminant or factoring.
• Factoring Sum/Difference: Techniques like those seen in the exercise when equations are summed or subtracted can uncover relationships between variables.

In exercise (d), adding equations gave insight into simplifying it into a recognizable quadratic form. By rewriting and factoring this form into \( (x+y)^2 = 4\), solutions could be simply understood for the sum of the variables.