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Magazine Chapter 5: Problem 68
Solve the system. $$ \begin{array}{l} 3^{x}-9^{y}=18 \\ 3^{x}+3^{y}=30 \end{array} $$
Short Answer
Expert verified
x = 3, y = 1
Step by step solution
01
Rewrite the Equations
Rewrite the equations in terms of powers of 3 to make them easier to solve. Recall that \[ 9 = 3^2 \], so the system of equations can be rewritten as:\[ \begin{array}{l} 3^{x} - (3^2)^{y} = 18 \ 3^{x} + 3^{y} = 30 \ \rightarrow 3^{x} - 3^{2y} = 18 \ \rightarrow 3^{x} + 3^{y} = 30 \end{array} \]
02
Substitute
Let \[ a = 3^x \] and \[ b = 3^y \]. Then the system becomes:\[ \begin{array}{l} a - b^2 = 18 \ a + b = 30 \ \rightarrow a - b^2 = 18 \ \rightarrow a + b = 30 \end{array} \]
03
Isolate One Variable
From the second equation, solve for \[ a \]: \[ a = 30 - b \]
04
Substitute into First Equation
Substitute \[ a = 30 - b \] into the first equation: \[ (30 - b) - b^2 = 18 \]
05
Simplify and Solve for b
Simplify the equation: \[ 30 - b - b^2 = 18 \rightarrow -b^2 - b + 12 = 0 \rightarrow b^2 + b - 12 = 0 \]. Factor the quadratic equation:\[ (b + 4)(b - 3) = 0 \]. Thus, \[ b = -4 \] or \[ b = 3 \]. Since \[ b = 3^y \], \[ b \] must be positive, so \[ b = 3 \].
06
Solve for a
Substitute \[ b = 3 \] back into \[ a = 30 - b \]: \[ a = 30 - 3 \rightarrow a = 27 \]
07
Determine x and y
Recall \[ a = 3^x \] and \[ b = 3^y \]. Thus:\[ 27 = 3^x \] and \[ 3 = 3^y \]. Solving for \[ x \] and \[ y \]:\[ 3^x = 3^3 \rightarrow x = 3 \]\[ 3^y = 3^1 \rightarrow y = 1 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
exponential equations
An exponential equation is one in which the variable appears in the exponent. For example, in the equation \(3^x + 3^y = 30\), the variables x and y are in the exponents. These equations often involve exponential functions, which are functions of the form \(a^x\), where a is a constant. To solve exponential equations, it is useful to express all the terms with the same base, as shown in the original exercise where 9 is rewritten as 3 squared.
substitution method
The substitution method is a technique to solve a system of equations by solving one equation for one variable and then substituting that expression into the other equation. This method simplifies the system step-by-step. In the given exercise, we substituted \(a = 3^x\) and \(b = 3^y\), which allowed us to transform the exponential equations into simpler algebraic ones. By solving \(a = 30 - b\) and substituting it into the first equation, we reduced the system to a single quadratic equation in terms of b.
quadratic equations
Quadratic equations are of the form \(ax^2 + bx + c = 0\). They are second-degree polynomial equations and can be solved in multiple ways, including factoring, using the quadratic formula, or completing the square. In the exercise, after substituting and simplifying the equations, we obtained a quadratic equation in terms of b, \(b^2 + b - 12 = 0\). This equation was then factored into \( (b + 4)(b - 3) = 0\), giving the solutions \(b = -4 \) and \( b = 3 \).
factoring
Factoring is the process of breaking down a complex expression into simpler factors that, when multiplied together, produce the original expression. It's a fundamental algebra technique used to solve quadratic equations. In the solution, the quadratic equation \(b^2 + b - 12 = 0\) was factored as \( (b + 4)(b - 3) = 0\). This step enabled us to find the roots of the equation, which are b = -4 and b = 3. Since b represents an exponential term, it must be positive, thus we only consider b = 3.
