Solving Exponential Equations A Step-by-Step Guide For 4^(x+10) = 17^(-3x)

Exponential equations are a fundamental topic in mathematics, appearing in various fields such as calculus, physics, and engineering. Solving these equations often involves leveraging the properties of logarithms to isolate the variable. This article will delve into the step-by-step solution of the exponential equation $4^{x+10} = 17^{-3x}$, providing a comprehensive guide for readers to understand the underlying principles and techniques. The equation $4^{x+10} = 17^{-3x}$ presents a classic problem in algebra that requires a solid understanding of exponential functions and logarithms. Our primary goal is to isolate $x$ and find its value. This involves transforming the equation into a more manageable form by applying logarithmic properties. We'll start by taking the logarithm of both sides, which allows us to bring the exponents down as coefficients. Then, we'll use algebraic manipulation to group the terms containing $x$ and finally solve for $x$. This process not only provides the solution but also enhances one's problem-solving skills in dealing with exponential equations. The solution presented here is meticulous, ensuring that each step is clear and logically sound, making it accessible for students and enthusiasts alike. This method can be applied to various exponential equations, making it a valuable tool in mathematical problem-solving. Through detailed explanations, this article aims to clarify any confusion and reinforce the understanding of how to tackle similar problems effectively. Understanding how to solve exponential equations is crucial for anyone studying advanced mathematics or pursuing a career in STEM fields.

We are given the exponential equation:

$4^{x+10} = 17^{-3x}$

Our task is to find the value of $x$ that satisfies this equation. This involves using properties of exponents and logarithms to isolate $x$. Solving this equation will demonstrate a practical application of logarithmic properties and algebraic manipulation. This equation is a quintessential example of how exponential functions and logarithms are intertwined in problem-solving. The equation might seem daunting at first, but with a methodical approach, it can be broken down into manageable steps. The key lies in understanding how logarithms can simplify exponential equations. By applying the logarithm to both sides, we can transform the exponents into coefficients, thus turning the problem into a linear equation. This linear equation is much easier to solve, and it ultimately leads us to the value of $x$. Moreover, solving such equations reinforces the understanding of mathematical principles and techniques that are widely used in various scientific and engineering disciplines. Therefore, mastering the method to solve this equation is not just about getting the answer but also about building a strong foundation for more advanced mathematical concepts. The skills learned here can be transferred to various problem-solving scenarios, making it an invaluable exercise in mathematical proficiency.

Step 1: Apply Logarithms to Both Sides

To solve for $x$, we first take the natural logarithm (ln) of both sides of the equation. This allows us to use the property of logarithms that $\ln(a^b) = b \ln(a)$.

$\ln(4^{x+10}) = \ln(17^{-3x})$

Applying the logarithmic power rule, we get:

$(x+10) \ln(4) = -3x \ln(17)$

This step is critical because it transforms the exponential equation into a linear equation, which is much easier to solve. The logarithm function essentially undoes the exponentiation, allowing us to bring the variables down from the exponent and work with them as coefficients. This transformation is a cornerstone technique in solving exponential equations and highlights the inverse relationship between exponential and logarithmic functions. By taking the logarithm of both sides, we maintain the equality while simplifying the equation's structure. It is important to choose a logarithm base that is convenient for the problem; here, the natural logarithm (base $e$) is used, but any base would work in principle. The choice of base often comes down to personal preference or the availability of tools like calculators that typically have natural and common logarithm functions readily available. The simplicity of this step belies its profound impact on the solution process, as it sets the stage for the subsequent algebraic manipulations that will lead us to the value of $x$.

Step 2: Expand and Rearrange the Equation

Next, we expand the left side of the equation:

$x \ln(4) + 10 \ln(4) = -3x \ln(17)$

Now, we want to group the terms containing $x$ on one side of the equation. Add $3x \ln(17)$ to both sides:

$x \ln(4) + 3x \ln(17) = -10 \ln(4)$

This rearrangement is a standard algebraic technique used to isolate the variable we are solving for. By moving all terms involving $x$ to one side, we create a structure that allows us to factor out $x$ and simplify the equation further. This step demonstrates the importance of algebraic manipulation in solving mathematical equations. The ability to rearrange terms without changing the equation's fundamental truth is a crucial skill in algebra. It allows us to transform complex equations into simpler forms that are easier to analyze and solve. In this case, by grouping the $x$ terms, we pave the way for factoring, which is the next logical step in isolating $x$. This process is not just about moving symbols around; it's about understanding the underlying mathematical relationships and using them to our advantage. The careful and methodical rearrangement of terms is a hallmark of effective problem-solving in mathematics, and this step exemplifies that principle.

Step 3: Factor out $x$

Factor out $x$ from the left side:

$x(\ln(4) + 3 \ln(17)) = -10 \ln(4)$

This step is a crucial algebraic manipulation that simplifies the equation by isolating $x$ as a common factor. Factoring is a fundamental technique in algebra, and its application here allows us to reduce the complexity of the equation. By factoring out $x$, we transform the sum of two terms involving $x$ into a single term, making it easier to solve for $x$. This step highlights the importance of recognizing common factors in algebraic expressions and using them to simplify equations. Factoring is not just a mechanical process; it requires an understanding of the structure of the equation and how the distributive property works in reverse. This skill is essential for solving a wide range of algebraic problems, and its mastery enhances one's ability to manipulate equations effectively. The factored form of the equation is now much closer to a solution, as it clearly shows $x$ multiplied by a constant term, which can be easily isolated by division. This step is a testament to the power of algebraic techniques in simplifying mathematical problems and making them more tractable.

Step 4: Isolate $x$

To isolate $x$, divide both sides by $(\ln(4) + 3 \ln(17))$:

$x = \frac{-10 \ln(4)}{\ln(4) + 3 \ln(17)}$

This step is the culmination of our algebraic efforts, where we finally isolate $x$ and express it in terms of known logarithmic values. Dividing both sides of the equation by the term multiplying $x$ is a straightforward yet critical step in solving for $x$. This action undoes the multiplication, leaving $x$ alone on one side of the equation. The resulting expression for $x$ is a precise solution, although it may not be immediately interpretable as a numerical value. It is important to note that this step assumes that the denominator $(\ln(4) + 3 \ln(17))$ is not zero, which is clearly the case here since both $\ln(4)$ and $\ln(17)$ are positive. This step underscores the importance of understanding the properties of algebraic operations and how they can be used to isolate variables. The ability to perform this kind of manipulation is a cornerstone of algebraic problem-solving and is essential for anyone working with mathematical equations. The isolated expression for $x$ is a significant achievement, as it represents the solution to the original exponential equation.

Step 5: Simplify the Expression (Optional)

We can further simplify the expression using logarithm properties. Recall that $n \ln(a) = \ln(a^n)$. Thus, $3 \ln(17) = \ln(17^3)$.

$x = \frac{-10 \ln(4)}{\ln(4) + \ln(17^3)}$

Also, $\ln(a) + \ln(b) = \ln(ab)$, so $\ln(4) + \ln(17^3) = \ln(4 \cdot 17^3)$.

$x = \frac{-10 \ln(4)}{\ln(4 \cdot 17^3)}$

While not strictly necessary, simplifying the expression can make it more elegant and easier to evaluate numerically. This step demonstrates the power of logarithmic properties in manipulating and simplifying expressions. By applying the rules for logarithms of powers and sums, we transform the denominator into a single logarithm, making the entire expression more compact. This simplification is not just about aesthetics; it can also make the expression easier to work with in further calculations or analyses. The ability to manipulate logarithmic expressions using these properties is a valuable skill in mathematics and its applications. It allows us to rewrite expressions in more convenient forms, which can be crucial in solving equations or evaluating functions. The simplified expression for $x$ is a testament to the elegance and efficiency of logarithmic transformations.

Step 6: Approximate the Value of $x$

Using a calculator, we can approximate the value of $x$:

$x ≈ \frac{-10 \cdot 1.3863}{\ln(4 \cdot 17^3)} ≈ \frac{-13.863}{\ln(19652)} ≈ \frac{-13.863}{9.885} ≈ -1.402$

Thus, $x ≈ -1.402$.

This final step provides a numerical approximation of the solution, making it more tangible and interpretable. While the exact solution is expressed in terms of logarithms, the approximate value gives us a concrete number that we can use in practical applications. This step highlights the importance of numerical methods in mathematics, as they allow us to translate abstract expressions into real-world values. The use of a calculator or computational tool is often necessary to obtain these approximations, and it is crucial to understand how to use these tools effectively. The approximate value of $x$ provides a useful check on the solution, as it allows us to verify that the answer is reasonable in the context of the original equation. Moreover, this numerical solution can be used in further analyses or applications where a precise value is needed. The process of approximating the solution underscores the interplay between symbolic and numerical methods in mathematical problem-solving.

The solution to the equation $4^{x+10} = 17^{-3x}$ is:

$x = \frac{-10 \ln(4)}{\ln(4) + 3 \ln(17)} ≈ -1.402$

This final answer encapsulates the entire solution process, providing both the exact form in terms of logarithms and the approximate numerical value. Presenting the solution in this way ensures clarity and completeness, allowing the reader to understand both the symbolic representation and the practical magnitude of the answer. The exact solution is valuable for theoretical purposes, while the approximate value is useful for applications and interpretations in real-world contexts. This comprehensive solution demonstrates the power of algebraic and logarithmic techniques in solving exponential equations. It also highlights the importance of a systematic approach to problem-solving, where each step is logically justified and contributes to the final result. The process of solving this equation reinforces the understanding of exponential functions, logarithms, and their properties, which are fundamental concepts in mathematics and its applications. The final answer not only solves the specific problem but also serves as a template for tackling similar exponential equations in the future. By understanding the underlying principles and techniques, one can confidently approach a wide range of mathematical challenges.