Problem 67 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, solve for \(x\) or \(t\). $$ 5^{2 x}=15 $$

Step-by-Step Solution

Answer

The solution is \(x \approx 0.682606\).

Step 1: Rewrite the equation in logarithmic form

An exponential equation \(a^b = c\) can be rewritten in logarithmic form as \(\log_a c = b\). Implementing this, the equation transforms to \(\log_5 15 = 2x\).

Step 2: Solve for the variable \(x\)

To isolate \(x\), divide both sides of the equation by 2. Thus, we result in \(x = \frac{1}{2} \log_5 15\).

Step 3: Express the solution in decimal form

Entering \(\frac{1}{2} \log_5 15\) into a scientific calculator produces a decimal value for \(x\).

Key Concepts

Logarithmic Form

When working with exponential equations, it's often useful to change them into logarithmic form. This transformation is essential because it makes the equation easier to solve, especially when the variable is in the exponent. Consider the exponential equation, \(a^b = c\). By rewriting this in logarithmic form, we have \(\log_a c = b\). This transformation essentially asks, "to what power should we raise \(a\) to get \(c\)?" For the exercise at hand, the equation \(5^{2x} = 15\) becomes \(\log_5 15 = 2x\). By converting an exponential equation to its logarithmic form, it becomes easier to handle and solve for the unknown variable, which in this case is \(x\).
Understanding the concept of logarithmic form is crucial for solving equations where the unknown is in an exponent. It helps in simplifying the equation, thereby making the process of isolating and solving for the variable more straightforward.

Solving for Variables

Once the equation is transformed into logarithmic form, the next step is solving for the variable, which is often the primary goal when dealing with mathematical equations. Let's look at \(\log_5 15 = 2x\). Since we want to find \(x\), we need to isolate it on one side of the equation. By dividing both sides by 2, the equation becomes \(x = \frac{1}{2} \log_5 15\).
Solving for a variable involves using algebraic manipulations to make the variable the subject of the formula. In our example, by dividing by 2, we effectively separate \(x\) and simplify our expression, making it ready for further calculations. This approach can be applied to various mathematical scenarios and is part of the fundamental skill set needed for problem-solving in math. The goal is always to get the variable on one side and all numbers and operations on the other.

Scientific Calculator Usage

After isolating the variable, it's time to compute the exact numerical value. This is where a scientific calculator becomes invaluable. Scientific calculators can handle more complex math functions like logarithms, roots, powers, and more, that are not possible or easy with just simple arithmetic.
In this situation, we need to compute \(\frac{1}{2} \log_5 15\). Since most scientific calculators are based on the common logarithm (base 10) or natural logarithm (base \(e\)), you may need to use the change of base formula: \(\log_5 15 = \frac{\log_{10} 15}{\log_{10} 5}\).

• First, calculate \(\log_{10} 15\).
• Next, calculate \(\log_{10} 5\).
• Divide the two results to find \(\log_5 15\).
• Finally, multiply the result by \(\frac{1}{2}\) to find \(x\).

Scientific calculators make it simpler to compute these values, providing an accurate and efficient means to solve equations. Understanding how to use them is a vital skill for tackling advanced mathematical problems effectively and accurately.