Problem 32 - Algebra 2 and Trigonometry

Question

Solve each equation and check. \((0.01)^{2 x}=100^{2-x}\)

Step-by-Step Solution

Answer

The solution is \(x = -2\).

Step 1: Rewrite Constants as Powers of 10

First, recognize that both 0.01 and 100 can be written as powers of 10. - 0.01 is equivalent to \(10^{-2}\) because it is \(1/100\).- 100 is equivalent to \(10^2\) because it is \(10 \times 10\).So rewrite the equation as:\[ (10^{-2})^{2x} = (10^2)^{2-x} \]

Step 2: Apply the Power of a Power Rule

Use the property of exponents that states \((a^m)^n = a^{m \cdot n}\) to simplify both sides of the equation.- The left side becomes \((10^{-2})^{2x} = 10^{-4x}\).- The right side becomes \((10^2)^{2-x} = 10^{4 - 2x}\).Now, the equation is:\[ 10^{-4x} = 10^{4-2x} \]

Step 3: Equate the Exponents

Since the bases are the same (both are 10), set the exponents equal to each other:\[-4x = 4 - 2x\]

Step 4: Solve for x

Solve the equation from Step 3:1. Add \(2x\) to both sides to get: \[-4x + 2x = 4 - 2x + 2x\] \[-2x = 4\]2. Divide both sides by -2: \[x = -2\]

Step 5: Verify the Solution

Substitute \(x = -2\) back into the original equation to check:- Original equation: \((0.01)^{2x} = 100^{2-x}\)- Substitute \(x = -2\): \[(0.01)^{2(-2)} = 100^{2+2}\] \[(0.01)^{-4} = 100^4\]- Compute both sides: \((0.01)^{-4} = (10^{-2})^{-4} = 10^8\), and \(100^4 = (10^2)^4 = 10^8\).- Both sides equal \(10^8\), confirming that \(x = -2\) is correct.

Key Concepts

Equation Solving

Solving equations is like unraveling a puzzle. You have an equation, which is a mathematical statement showing that two expressions are equal, and your task is to find the value of the variable that makes this statement true. In any equation solving process, the first step is understanding the equation's structure, the operations involved, and identifying what you are solving for.
To solve an equation, you typically manipulate it using mathematical operations until you isolate the variable on one side. This means converting an equation into a statement where the unknown is equal to a specific number. Common techniques include adding or subtracting terms from both sides of the equation, dividing or multiplying each side by a number, and simplifying expressions.
The goal is to systematically simplify the equation following arithmetic rules until the variable is isolated and you uncover its value. Being methodical and consistent with each step ensures that the solution is correct.

Exponents

Exponents are a shorthand way to express repeated multiplication of the same number. For example,

• \(10^2\) means \(10\) multiplied by itself, which equals \(100\).
• Similarly, \(a^3\) indicates that the base \(a\) is multiplied three times: \(a \times a \times a\).

Exponents play a crucial role in algebra, mainly when dealing with large numbers, as they provide a compact way of expressing these values. Understanding how to manipulate exponents is key to simplifying algebraic expressions and solving equations.
For instance, when multiplying like bases, you add the exponents: \(a^m \times a^n = a^{m+n}\). It is just one of several rules that make working with exponents more manageable. By mastering these rules, solving complex algebraic problems becomes more straightforward.

Power of a Power Rule

The power of a power rule is essential for simplifying expressions involving exponents raised to another power. This rule states that

• \((a^m)^n = a^{m \cdot n}\).

It means that when an exponent is raised to another exponent, you multiply the exponents.
For instance, in the equation

• \((10^{-2})^{2x}\), applying the power of a power rule results in \(10^{-4x}\).
• Similarly, \((10^2)^{2-x}\) becomes \(10^{4-2x}\).

The power of a power rule is instrumental when simplifying equations since it allows you to express complex exponent forms in a simpler, more manageable way.
Understanding this rule enables you to transform and resolve problems more effectively, especially when dealing with equations involving exponential expressions.

Verifying Solutions

Verifying solutions is the final step in confirming if the solution you've obtained for an equation is correct. It involves substituting the found value back into the original equation to check if both sides of the equation still hold true.
In our exercise, once we solved for \(x = -2\), it’s crucial to verify this by plugging \(x\) back into the original equation:

• \((0.01)^{2 \times -2} = 100^{2 - (-2)}\)
• The left simplifies as \((0.01)^{-4}\) equivalent to \(10^{8}\).
• The right simplifies to \(100^4\) equivalent to \(10^{8}\).

Both sides equate to \(10^8\). Therefore, the solution \(x = -2\) is indeed correct. Verifying solutions like this ensures accuracy and builds confidence in algebraic problem-solving.