Solve For X In 5^(3x) = 13: A Step-by-Step Guide

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In the realm of mathematics, solving exponential equations is a fundamental skill that bridges algebra and calculus. Exponential equations, where the variable appears in the exponent, often require logarithmic techniques to isolate the variable. In this comprehensive guide, we will tackle the equation 53x=13{ 5^{3x} = 13 }, providing a step-by-step solution and illuminating the underlying principles.

Understanding Exponential Equations

Before diving into the solution, it's crucial to grasp the essence of exponential equations. An exponential equation is an equation in which the variable appears in the exponent. These equations arise in various contexts, including population growth, radioactive decay, and compound interest. The key to solving them lies in understanding the inverse relationship between exponential and logarithmic functions.

The equation 53x=13{ 5^{3x} = 13 } exemplifies this type of problem. Our goal is to find the value of x{ x } that satisfies this equation. The presence of x{ x } in the exponent necessitates the use of logarithms to bring it down and isolate it.

To effectively solve for x{ x }, we need to utilize logarithms, which are the inverse functions of exponentials. The logarithm of a number to a given base is the exponent to which the base must be raised to produce that number. In simpler terms, if ab=c{ a^b = c }, then logac=b{ \log_a{c} = b }. This inverse relationship is the cornerstone of solving exponential equations. Understanding the properties of logarithms, such as the power rule, which states that logbac=clogba{ \log_b{a^c} = c \log_b{a} }, will be crucial in our solution. This property allows us to bring the exponent down, transforming the exponential equation into a more manageable algebraic form. The careful application of logarithmic properties, combined with algebraic manipulation, will allow us to isolate x{ x } and determine its value. The process of solving 53x=13{ 5^{3x} = 13 } demonstrates a blend of algebraic technique and logarithmic insight, highlighting the interconnectedness of different areas within mathematics. By mastering these techniques, one gains a deeper understanding of mathematical relationships and problem-solving strategies applicable across various disciplines. Understanding these concepts is critical for success in advanced mathematics and its applications in the real world.

Step-by-Step Solution

1. Apply Logarithms to Both Sides

The initial step in solving 53x=13{ 5^{3x} = 13 } is to apply a logarithm to both sides of the equation. The choice of the logarithmic base is flexible, but the common logarithm (base 10) or the natural logarithm (base e{ e }) are frequently used due to their availability on calculators. For this solution, we'll use the natural logarithm (ln{ \ln }), which is the logarithm to the base e{ e }. Applying the natural logarithm to both sides, we get:

ln(53x)=ln(13){ \ln(5^{3x}) = \ln(13) }

This step is critical because it allows us to utilize the properties of logarithms to simplify the equation. By taking the logarithm of both sides, we maintain the equality while introducing a function that will help us isolate the variable x{ x }. The next step will involve using the power rule of logarithms to bring the exponent down, further simplifying the equation and moving us closer to a solution. Remember, the key to solving exponential equations is to transform them into a form where the variable can be easily isolated, and logarithms are the perfect tool for this transformation.

2. Use the Power Rule of Logarithms

The power rule of logarithms states that logb(ac)=clogb(a){ \log_b(a^c) = c \log_b(a) }. This rule is essential for simplifying logarithmic expressions where the argument of the logarithm is raised to a power. Applying this rule to the left side of our equation, ln(53x)=ln(13){ \ln(5^{3x}) = \ln(13) }, we can bring the exponent 3x{ 3x } down as a coefficient:

3xln(5)=ln(13){ 3x \ln(5) = \ln(13) }

This transformation is a significant step in solving for x{ x }. By using the power rule, we've effectively converted the exponential equation into a linear equation in terms of x{ x }. This makes the equation much easier to manipulate algebraically. The next step will involve isolating x{ x } by dividing both sides of the equation by the coefficient of x{ x }. This step demonstrates the power of logarithmic properties in simplifying complex equations and making them solvable through basic algebraic techniques. The ability to apply the power rule correctly is a fundamental skill in solving exponential and logarithmic equations, and it is crucial for success in more advanced mathematical problems.

3. Isolate x

Now that we have the equation 3xln(5)=ln(13){ 3x \ln(5) = \ln(13) }, the next step is to isolate x{ x }. To do this, we need to divide both sides of the equation by 3ln(5){ 3 \ln(5) }. This operation will leave x{ x } by itself on the left side, giving us the solution in terms of logarithms:

x=ln(13)3ln(5){ x = \frac{\ln(13)}{3 \ln(5)} }

This is the exact solution for x{ x }. It's expressed in terms of natural logarithms, which is a precise and accurate representation. However, for practical applications, we often need a numerical approximation. In the next step, we'll use a calculator to find the approximate decimal value of x{ x }. Isolating the variable is a crucial step in solving any equation, and in this case, it demonstrates the effectiveness of using logarithmic properties to transform and simplify the original exponential equation. The ability to manipulate equations and isolate variables is a fundamental skill in mathematics and is essential for problem-solving in various fields.

4. Calculate the Numerical Value

To find the numerical value of x{ x }, we can use a calculator to evaluate the expression x=ln(13)3ln(5){ x = \frac{\ln(13)}{3 \ln(5)} }. The natural logarithm function (ln{ \ln }) is available on most scientific calculators. Inputting the expression into a calculator, we get:

x2.56493×1.60942.56494.82820.5313{ x \approx \frac{2.5649}{3 \times 1.6094} \approx \frac{2.5649}{4.8282} \approx 0.5313 }

Therefore, the approximate value of x{ x } is 0.5313. This value satisfies the original equation 53x=13{ 5^{3x} = 13 }. It's important to note that this is an approximate value, as we've rounded the result to four decimal places. In many practical applications, an approximate solution is sufficient, but it's always good to keep the exact solution in terms of logarithms for maximum precision. This step highlights the importance of using calculators or computational tools to obtain numerical solutions from mathematical expressions. It also underscores the distinction between exact solutions, which are expressed in terms of mathematical constants and functions, and approximate solutions, which are numerical values obtained through computation.

Verification

To ensure our solution is correct, we can substitute the approximate value of x{ x } back into the original equation and see if it holds true. Substituting x0.5313{ x \approx 0.5313 } into 53x=13{ 5^{3x} = 13 }, we get:

53×0.5313=51.593912.99913{ 5^{3 \times 0.5313} = 5^{1.5939} \approx 12.999 \approx 13 }

Since the result is approximately 13, our solution is verified. The slight discrepancy is due to rounding errors in our approximation. This step is crucial in the problem-solving process as it provides a check for the accuracy of our solution. By substituting the calculated value back into the original equation, we can confirm whether our steps and calculations were correct. Verification not only ensures the correctness of the solution but also enhances our understanding of the problem and the solution process. It reinforces the importance of precision and accuracy in mathematical calculations and highlights the potential impact of rounding errors in approximate solutions. This practice is valuable in various mathematical contexts and helps build confidence in one's problem-solving abilities.

Conclusion

In this comprehensive guide, we've demonstrated how to solve the exponential equation 53x=13{ 5^{3x} = 13 } step by step. By applying logarithms, using the power rule, isolating the variable, and calculating the numerical value, we found that x0.5313{ x \approx 0.5313 }. This process showcases the power of logarithms in solving exponential equations and provides a solid foundation for tackling more complex problems in mathematics and related fields. The ability to solve exponential equations is a valuable skill in many areas, including science, engineering, and finance.

The key takeaways from this solution include the importance of understanding the inverse relationship between exponential and logarithmic functions, the application of logarithmic properties, and the use of algebraic manipulation to isolate variables. Furthermore, we emphasized the significance of verifying solutions to ensure accuracy. Mastering these techniques will empower you to confidently solve a wide range of exponential equations and apply these skills in various practical contexts. This detailed walkthrough not only provides a solution to a specific problem but also equips the reader with a broader understanding of the principles and techniques involved in solving exponential equations, fostering a deeper appreciation for the elegance and power of mathematics.