Problem 38 - Algebra 2

Question

If \(2^{r+5}=2^{2 r-1},\) what is the value of \(r ?\)

Step-by-Step Solution

Answer

The value of \(r\) is 6.

Step 1: Equating the Exponents

Since the base of both exponentials is the same (2), we can equate the exponents according to the property that if \(a^m = a^n\), then \(m = n\). Therefore, we have \(r+5=2r-1.\)

Step 2: Isolating the Variable

Subtract \(r\) from both sides to simplify the equation: \(5 = r - 1.\)

Step 3: Solving for r

Add 1 to both sides of the equation to find the value of \(r\): \(5 + 1 = r\). This simplifies to \(r = 6.\)

Key Concepts

Equating Exponents

When dealing with exponent problems, an important rule is that if you have the same base on both sides of an equation, you can equate the exponents. This is because exponential functions with the same base are one-to-one functions. So if you have something like \(a^m = a^n\), it implies that \(m = n\). This makes it much easier to solve equations since you only need to focus on the exponents without worrying about the base.

• In the problem \(2^{r+5} = 2^{2r-1}\), both sides have the same base, which is 2.
• By applying the rule of equating exponents, we directly write \(r + 5 = 2r - 1\).
• This step reduces the complexity and paves the way for solving the equation efficiently.

Understanding this principle is crucial as it is a fundamental tool in algebra that simplifies exponential equations, allowing you to solve for the unknown variable.

Solving Equations

Once you have equated the exponents, the next step is to solve for the variable. Solving equations is a basic algebraic skill that involves finding the value of the unknown variable that makes the equation true.

• After equating the exponents, we have the equation \(r + 5 = 2r - 1\).
• This equation involves simple linear terms of the variable \(r\).
• The goal is to isolate the variable on one side of the equation.

In our scenario, we start by subtracting \(r\) from both sides:\[5 = 2r - r - 1\]This simplification now shows us the equation in simpler terms, making it linear and easy to manipulate further. Always remember that when solving equations, each step must maintain the equality.

Algebraic Manipulation

Algebraic manipulation is all about rearranging equations to make them easier to work with. It's a key part of solving equations and is done through operations such as adding, subtracting, multiplying, or dividing terms on both sides of the equation.

• Continuing from \(5 = r - 1\), we need to find \(r\).
• We add 1 to both sides to simplify the equation further.
• This gives \(5 + 1 = r\), resulting in \(r = 6\).

These steps highlight the importance of algebraic manipulation in isolating the variable. Each operation is carefully chosen to ensure that the variable is increasingly isolated, ultimately allowing for the straightforward identification of its value. Mastery of these skills is essential for tackling more complex algebraic problems in the future.