Problem 31 - Algebra A Combined Function

Question

Solve. $$ 27^{x+1}=9 $$

Step-by-Step Solution

Answer

The solution is \( x = -\frac{1}{3} \).

Step 1: Understand the Problem

The equation to solve is \( 27^{x+1} = 9 \). We need to find the value of \( x \) that satisfies this equation.

Step 2: Express Both Sides with Common Base

Observe that both 27 and 9 can be expressed as powers of the same number. 27 is \( 3^3 \) and 9 is \( 3^2 \). Rewriting both sides, the equation becomes \( (3^3)^{x+1} = 3^2 \).

Step 3: Apply the Power of a Power Rule

Apply the power of a power rule \( (a^m)^n = a^{m imes n} \). The left side becomes \( 3^{3(x+1)} = 3^2 \).

Step 4: Set Exponents Equal

Now that the bases are the same, we can set the exponents equal to each other: \( 3(x+1) = 2 \).

Step 5: Solve for x

Solve the equation \( 3(x+1) = 2 \). First, distribute the 3: \( 3x + 3 = 2 \). Then, subtract 3 from both sides to get \( 3x = -1 \). Finally, divide by 3 to find \( x = -\frac{1}{3} \).

Key Concepts

Exponent Rules

When dealing with exponential equations, a solid grasp of exponent rules is essential. One fundamental rule used in the original exercise solution is the power of a power rule, which states that

• \( (a^m)^n = a^{m \times n} \)

This rule makes it easier to simplify complex exponential expressions.
You apply this by multiplying the exponents when raising a power to another power. For instance, if you encounter \((3^3)^{x+1}\), you multiply the exponent 3 by the expression \(x+1\), resulting in\(3^{3(x+1)}\).
Another important rule is how equating exponents works. Once both sides of an equation have the same base, their exponents can be set equal to solve for the unknown variables. This is possible because of the property that if \(a^m = a^n\), then \(m = n\).
These foundational rules simplify complex equations, reducing them to manageable algebraic expressions.

Base Conversion

Base conversion is crucial when simplifying exponential equations. It's all about expressing numbers as powers of the same base. This process helps in revealing the relationship between the terms involved.
In our example, the numbers 27 and 9 might seem unrelated at first. However, by converting them to powers of a common base, the solution becomes straightforward.
Here's how it was done:

• 27 is rewritten as a power of 3, since \(27 = 3^3\).
• Similarly, 9 can be rewritten with the same base, where \(9 = 3^2\).

Now that both numbers are expressed as powers of 3, the task becomes easier. We can directly apply the exponent rules without any additional conversions, streamlining the process and leading to quicker solutions.

Equation Solving

The art of solving equations frequently involves simplifying and restructuring them to find unknown variables. Once the exponents have common bases, you can set them equal to each other.
From this point, it resembles solving a typical algebraic equation. Here's a breakdown using the example:

• After conversion and simplification, you're left with \(3(x+1) = 2\).
• The task is simple algebra: distribute and simplify this equation.
• First, distribute the 3 across the \(x+1\), yielding \(3x + 3\).
• Then, subtract 3 from both sides to isolate terms involving x.
• This results in \(3x = -1\).
• Finally, divide both sides by 3 to find the value of x, resulting in \(x = -\frac{1}{3}\).

Solving equations combines a systematic approach with arithmetic precision, allowing for solving for the unknowns.

Algebraic Manipulation

Algebraic manipulation is the backbone of problem-solving in algebra. It's about rearranging and simplifying expressions to make equations solvable.
This requires:

• Applying distributive properties, such as \(a(b+c) = ab + ac\).
• Moving terms across the equation by performing operations such as addition, subtraction, multiplication, or division.
• Isolating variables to one side, making the equation simpler to solve.

In the exercise, the expression \(3(x+1)\) was expanded to \(3x + 3\), demonstrating the use of distributive property.
Subtracting 3 from both sides to simplify \(3x = 2 - 3\) shows how terms can be moved across the equal sign efficiently. Finally, dividing by 3 completed the solution by isolating the variable x.
This mastery of algebraic manipulation is pivotal in crafting efficient solutions and tackling challenging equations.