91Ó°ÊÓ

Solving exponential equations often involves simplifying the expressions to make them more manageable. In this case, variable substitution is a powerful technique. When faced with an equation like \(e^{2x} - e^{x} - 6 = 0\), replacing parts of the equation with a new variable can simplify things significantly.

In this problem, we substitute \(y = e^x\). This substitution helps reduce the complexity because it transforms \(e^{2x}\) into \(y^2\). The original equation then turns into a quadratic form: \(y^2 - y - 6 = 0\).

This technique is very useful when dealing with exponential equations because it reduces them to a quadratic equation, which is something we know how to solve with standard methods. Always remember, the new variable you introduce must be reversible, meaning you should be able to express the original variable in terms of your new one later.

A quadratic equation is an equation that can be arranged in the standard form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents an unknown variable. They often appear in various fields like physics, engineering, and finance.

In our exercise, the equation \(y^2 - y - 6 = 0\) is a quadratic equation. Solving quadratic equations can be achieved through several methods such as factoring, using the quadratic formula, or completing the square.

• Factoring involves finding two numbers that multiply to the constant term (in this case, \(-6\)) and add up to the coefficient of the middle term (here, \(-1\)).
• Quadratic equations usually have two possible solutions, found by considering each factor separately.

For our equation, this factoring yields \((y - 3)(y + 2) = 0\), leading to two potential solutions: \(y = 3\) or \(y = -2\).
Once factored, the problem simplifies to solving for \(y\) by setting each factor equal to zero.

Once you solve the quadratic equation and determine the values of \(y\), you need to substitute back to find \(x\). This is where the natural logarithm comes into play.

The natural logarithm is the logarithm to the base \(e\), well known in mathematics as the irrational number approximately equal to 2.71828. It is denoted as \(\ln x\) and is the inverse of the exponential function.

Given \(y = e^{x}\), to solve for \(x\), apply the natural logarithm: \(x = \ln(y)\).

• For \(y = 3\), take the natural logarithm on both sides to get \(x = \ln(3)\).
• The solution \(y = -2\) doesn’t work because \(e^{x}\) (an exponential function) is always positive. You can't take the natural log of a negative number in this context.

This understanding is crucial because it highlights the limitation of the exponential functions while also showing how natural logarithms serve as a critical tool in reverting your substitution, providing the means to solve for the original variable \(x\). Hence, the solution to the exponential equation is \(x = \ln(3)\).