Question

The roots of the equation \(5^{2 x}+(-30) 5^{x}+125=0\) are (a) \(\\{1,2\\}\) (b) \(\\{5,10\\}\) (c) \(\\{5,26\\}\) (d) \(\\{4,5\\}\)

Short answer

Answer: (a) \(\{1,2\}\)

Step-by-step solution

Substitution

Let \(y = 5^x\). The equation changes to \(y^2 - 30y + 125 = 0\). #Step 2: Solving the Quadratic Equation#

Solving the Quadratic Equation

Solve the equation \(y^2 - 30y + 125 = 0\). Using the quadratic formula, we get: \(y = \frac{-(-30) \pm \sqrt{(-30)^2 - 4\cdot1\cdot125}}{2\cdot1}=\frac{30 \pm \sqrt{900 - 500}}{2}=\frac{30 \pm \sqrt{400}}{2}=\frac{30 \pm 20}{2}\) We have two possible values for y: \(y_1 = \frac{30 + 20}{2}=25\) \(y_2 = \frac{30 - 20}{2}=5\) #Step 3: Reverse Substitution#

Reverse Substitution

Revert the substitution, we get \(5^x = y\). So, we have \(5^x = 25\) and \(5^x = 5\). Applying logarithm: For \(5^x = 25\), take log base 5 on both sides, we get \(x = \log_5{25} = 2\). For \(5^x = 5\), take log base 5 on both sides, we get \(x = \log_5{5} = 1\). So, the roots are x = 1, and 2. Therefore, the correct option is: (a) \(\{1,2\}\)

Key concepts

Quadratic Formula

Understanding the quadratic formula is essential when dealing with equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and `a` isn't zero. The formula itself is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). It's a powerful tool allowing us to find the roots of any quadratic equation by substituting the values of the coefficients.

Using the quadratic formula can sometimes lead to complex numbers when the discriminant (\( b^2 - 4ac \)) is negative, indicating that there are no real roots. However, in our exercise, the discriminant was positive, yielding two real roots. To apply the formula correctly, it's essential to perform each calculation step with care, ensuring the correct order of operations is followed.

This method helps us solve the modified equation \( y^2 - 30y + 125 = 0 \) efficiently, guiding us to the conclusion that the roots for \( y \) are 25 and 5. These roots play a crucial role when we substitute back to find the original variable \( x \) in our exercise.

Substitution Method

The substitution method is a powerful algebraic technique, especially useful when dealing with equations where the traditional approach might be cumbersome. This method involves replacing a term with a simpler expression or a variable that makes the equation easier to solve.

In the given exercise, we performed substitution by letting \( y = 5^x \), which transformed the original complex exponential equation into a simpler quadratic equation. The key advantage of using substitution is that it can convert a difficult or unfamiliar equation into a format that we are more equipped to solve, such as a quadratic equation. This simplifies the solving process and allows us to apply methods that we are comfortable with, such as the quadratic formula.

However, keep in mind that after solving for the new variable, reversing the substitution is a critical step. This ensures that we find the solution(s) for the original variable, which is ultimately what's required. It's important to substitute back carefully to avoid errors and to verify the solutions within the original equation if possible.

Logarithms

Logarithms are the inverse operations of exponentiation. They answer the question: to what exponent must we raise a certain base to obtain a given number? The logarithm of a number is written as \( \log_b(x) \), where \( b \) is the base, and \( x \) is the number we're taking the logarithm of.

For example, if we have \( b^y = x \), then by applying logarithms we have \( y = \log_b(x) \). In our exercise, after applying the substitution back to \( 5^x \), we ended up with \( 5^x = 25 \) and \( 5^x = 5 \). Taking logarithms to the base 5 of both sides gave us our solutions for \( x \) as 2 and 1, respectively.

Logarithms are particularly useful in solving exponential equations, where the variable is an exponent. They bring the variable down from the exponent, making it possible to solve for the variable directly. Understanding the properties of logarithms, such as the product, quotient, and power rules, can further simplify complex exponential equations into simpler forms that can be solved with basic algebra.