Problem 27 - Beginning and Intermediate Algebra

Question

Solve the exponential equation algebraically. Then check using a graphing calculator. $$3^{2 x-1}=5^{x}$$

Step-by-Step Solution

Answer

To solve the exponential equation \(3^{2x-1} = 5^x\) algebraically, we apply the natural logarithm to both sides and use logarithmic properties: \((2x-1)\ln(3) = x\ln(5)\) Solve for x to find: \(x = \frac{\ln(3)}{2\ln(3) - \ln(5)}\) Verify the solution using a graphing calculator and find that the x-coordinate of the intersection point matches the value of x found algebraically. Therefore, the solution is \(x = \frac{\ln(3)}{2\ln(3) - \ln(5)}\).

Step 1: Rewrite the equation using natural logarithms

To solve for x in the exponential equation \(3^{2x-1} = 5^x\), we can apply the natural logarithm (ln) to both sides. \( \ln(3^{2x-1}) = \ln(5^x) \)

Step 2: Use logarithmic properties to simplify

Recall the logarithmic property \(\ln(a^b) = b\ln(a)\). Apply this property to both sides of the equation. \((2x-1)\ln(3) = x\ln(5)\)

Step 3: Solve for x

Now we need to solve for x. Start by expanding the terms: \(2x\ln(3) - \ln(3) = x\ln(5)\) Move the x term from the right side to the left side: \(2x\ln(3) - x\ln(5) = \ln(3)\) Factor out x: \(x(2\ln(3) - \ln(5)) = \ln(3)\) Now isolate x by dividing both sides by \((2\ln(3) - \ln(5))\): \(x = \frac{\ln(3)}{2\ln(3) - \ln(5)}\)

Step 4: Check with a graphing calculator

Use a graphing calculator to graph the two exponential functions: 1. \(y = 3^{2x-1}\) 2. \(y = 5^x\) Locate where the two graphs intersect. The x-coordinate of the intersection point should match the value of x we found in step 3. This will confirm that our solution is correct. In this case, when you graph the two functions and find their intersection point, you'll see that the x-coordinate matches the value of x that we obtained algebraically. Thus, our solution is correct. The solution for the exponential equation \(3^{2x-1} = 5^x\) is: \(x = \frac{\ln(3)}{2\ln(3) - \ln(5)}\)

Key Concepts

Logarithms

Logarithms are powerful tools in mathematics that help us deal with exponential equations by transforming them into linear equations. This transformation makes it easier to handle equations that otherwise might seem complex. At the core, a logarithm answers the question: “To what exponent should we raise a certain base to get a specific number?” For instance, if we have the equation \(a^b = c\), then \(log_a(c)\) gives us \(b\). In practical terms, when solving equations such as \(3^{2x-1} = 5^x\), we apply logarithms to simplify these into forms that are easier to solve directly. Here's why this works:

• Applying the logarithm to both sides of an exponential equation helps to "bring down" the exponent, making the equation more manageable.
• Once each side is logged, properties like \( log(a^b) = b log(a)\) help separate variables from constants, simplifying the problem significantly.

The aim is to solve for \(x\) using these principles, and by transforming and rearranging equations, we can effectively determine its value.

Graphing Calculator

A graphing calculator is a technological tool that makes visual representation of functions straightforward, enhancing our understanding of their behavior. When it comes to solving exponential equations, graphing calculators can provide a second way to verify the accuracy of our algebraic solutions. Here’s how you can use a graphing calculator:

• First, enter the two separate functions into the graphing area. For the equation \(3^{2x-1} = 5^x\), you’d graph \(y = 3^{2x-1}\) and \(y = 5^x\).
• Once graphed, look at where the two lines intersect. This intersection shows you the value of \(x\) where both sides of the equation are equal, thus matching our algebraic solution.

Using a graphing calculator this way not only helps in confirming solutions but also gives you a visual insight into the nature of the functions involved, such as their growth rates and critical points.

Algebraic Solutions

Algebraic solutions involve manipulating mathematical expressions using a series of logical steps to find the values of unknown variables. In the context of exponential equations, algebraic solutions require special considerations due to the nature of the functions involved. Here's how this unfolds in an equation such as \(3^{2x-1} = 5^x\):

• First, use the property of logarithms to rewrite the exponential equation in a linear form. This makes it easier to isolate the variable \(x\).
• Second, follow through a series of algebraic manipulations to group like terms and solve for \(x\). Factors such as dividing by a coefficient, factoring expressions, and isolating \(x\) on one side are key techniques.
• Finally, translate these steps into a solution that gives you an exact numeric answer or formula for the variable of interest.

Algebraic solutions are precise because they rely on established mathematical rules. Solving using algebra not only tests your understanding of these rules but also provides a systematic approach to tackling more complex equations.