91Ó°ÊÓ

The substitution method is a valuable solving technique used for equations that can be cumbersome to tackle in their original form, particularly in polynomial equations. In the given problem, the equation features exponential terms, specifically with the base of 5 raised to powers involving the variable \(x\). A clever way to simplify this is to substitute a new variable for the exponential term. For example, let \(y = 5^x\). This transforms \(5^{2x}\) into \((5^x)^2\), which can be rewritten as \(y^2\). Now, the equation changes from \(5^{2x} + 3 \times 5^x = 28\) to \(y^2 + 3y = 28\).
By converting the equation into an easier-to-solve form, we can bring it down to a standard quadratic equation. We then solve the quadratic equation for \(y\) and subsequently find the values of \(x\) using the relationship \(y = 5^x\). This method can dramatically simplify solving complex equations, revealing an underlying structure that is much more manageable.

Quadratic equations are often solved using the quadratic formula, which provides a systematic way to find the solutions, or 'roots', of the equation. The quadratic formula is written as:

• \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Here, \(a\), \(b\), and \(c\) are coefficients from the standard form of a quadratic equation: \(ax^2 + bx + c = 0\). In this exercise, after substituting \(y = 5^x\), the equation becomes \(y^2 + 3y - 28 = 0\). Thus, \(a = 1\), \(b = 3\), \(c = -28\).
In applying the quadratic formula, the discriminant \(b^2 - 4ac\) plays a crucial role as it determines the nature of the roots:

• If the discriminant is positive, there are two distinct real roots.
• If it's zero, there's exactly one real root.
• If negative, the roots are complex and not real numbers.

For our equation, the discriminant is \(3^2 - 4 \times 1 \times (-28) = 121\), which is positive and a perfect square. Hence, the equation has two real solutions: \(y = 4\) and \(y = -14\). However, only \(y = 4\) is valid for \(5^x = y\), as \(5^x\) must be a positive number.

Exponential equations involve variables in the exponents, such as \(5^x = 4\) in our final step after the substitution process. To solve such equations, logarithms are typically used to isolate the variable. Logarithms act as the inverse operation to exponentiation. In the case where \(5^x = 4\), taking the logarithm of both sides allows us to bring the variable \(x\) down from the exponent:

• \(x \log(5) = \log(4)\)

Solving for \(x\) involves dividing both sides by \(\log(5)\), leading to:

• \(x = \frac{\log(4)}{\log(5)}\)

This gives an approximate decimal solution that can be calculated numerically. The process of solving such exponential equations by taking logarithms is straightforward once the equation is appropriately set up. It enables us to effectively find the solutions even when the base and the exponent make calculations non-trivial. Understanding this method is crucial whenever dealing with equations that incorporate exponential terms.