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Chapter 10: Problem 100

Solve. $$\begin{array}{l} e^{x}-e^{x+y}=0 \\ e^{y}-e^{x-y}=0 \end{array}$$

Short Answer

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Step by step solution

01

- Analyze the First Equation

Start with the first equation: : e^{x} - e^{x + y} = 0. Factor out the common term : e^{x} :. This gives: e^{x}(1 - e^{y}) = 0. This equation is satisfied if either : e^{x} = 0 or : 1 - e^{y} = 0. Since e^{x} = 0 has no real solutions, our only solution is: e^{y} = 1.
02

- Solve for y

From : e^{y} = 1, we find that : y = backwards evaluate_: backwards evaluate_0. Now substitute y = 0 into the second equation.
03

- Analyze the Second Equation

The second equation is - : e^{y} - e^{x-y} = 0. Substitute y = 0 back into this equation, giving: backward_slash_not e^{0} - e^{x-0} =0. This simplifies to 1 - e^{x} = 0, which implies : e^{x} = 1.
04

- Solve for x

From : e^{x} = 1, it follows that : x = 0. So we have found that : x = 0 and : finding y = 0. The complete solution set is ( : 0,0).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solving Equations
In mathematics, solving equations is about finding the values of variables that make an equation true. In this exercise, we dealt with a system of two exponential equations. The goal was to find the values for x and y that satisfy both equations at the same time. An equation like \(e^{x} - e^{x + y} = 0\) appears complex, but by breaking it down step-by-step and factoring out common terms, solving becomes more manageable.
Exponential Functions
Exponential functions involve variables appearing in the exponent. For example, in the equation \(e^{x} - e^{x + y} = 0\), both terms include the number \(e\) raised to different powers. The function \(e^x\) grows very fast as x increases. In our solution, recognizing properties of exponential functions was key in simplifying and solving the equation. Exponential equations may have special properties, such as \(e^{0} = 1\). This insight helped us simplify and solve each part of the equations.
Factoring
Factoring is a method used to simplify expressions and solve equations. It involves expressing a complex expression as a product of simpler factors. In the first equation, \(e^{x} - e^{x + y} = 0\), we factored out the common term \(e^{x}\), transforming it into \(e^{x}(1 - e^{y}) = 0\). This allowed us to set each factor to zero separately and find potential solutions. Another example is understanding that \(1 - e^y = 0\) when \(e^x\) cannot be zero. Factoring helps in isolating variables and simplifying the problem incrementally.

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Most popular questions from this chapter

Find an equation of the parabola with a vertical axis of symmetry and vertex \((-1,2)\) and containing the point \((-3,1)\)

Convert to degree measure. $$\frac{\pi}{3}$$

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Convert to radian measure. $$120^{\circ}$$

Fill in the blank with the correct term. Some of the given choices will not be used. $$\begin{array}{ll}\text { piecewise function } & \text { ellipse }\\\ \text { linear equation } & \text { midpoint } \\ \text { factor } & \text { distance } \\ \text { remainder } & \text { one real-number } \\ \text { solution } & \text { solution } \\ \text { zero } & \text { two different real-number } \\\ x \text { -intercept } & \text { solutions } \\ y \text { -intercept } & \text { two different imaginary- } \\ \text { parabola } & \text { number solutions } \\ \text { circle } & \end{array}$$ For a quadratic equation \(a x^{2}+b x+c=0,\) if \(b^{2}-4 a c>0,\) the equation has ___________________ .
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