Solving Logarithmic Equations A Detailed Explanation Of Elsa's Problem

Elsa faces a fascinating challenge involving logarithmic equations, a cornerstone of algebraic mathematics. Her claim that the equation $\log _2 x=\log _2(3 x+5)+4$ has no solution invites us to explore the depths of logarithmic properties and equation-solving techniques. In this comprehensive analysis, we'll dissect the equation, apply relevant logarithmic rules, and rigorously determine whether Elsa's assertion holds true. Understanding the intricacies of logarithmic equations is crucial not only for academic success but also for various real-world applications, from computer science to engineering. Thus, our journey into this problem will be both enlightening and practically valuable.

Logarithmic Equations are a unique breed within algebra, demanding careful attention to their inherent properties and restrictions. Elsa's equation, $\log _2 x=\log _2(3 x+5)+4$, immediately presents us with a fascinating puzzle. To begin, it's essential to remember that the logarithm function, particularly $\log _2 x$, is only defined for positive values of $x$. This is a crucial constraint we must bear in mind throughout the solving process. The argument inside the logarithm must be strictly greater than zero because logarithms essentially ask, "To what power must we raise the base (in this case, 2) to obtain this number?" You cannot raise a positive number to any power and get zero or a negative number.

Now, let's dive into how we might approach solving this equation. The first step typically involves isolating the logarithmic terms and attempting to combine them. We can start by subtracting $\log _2(3x + 5)$ from both sides of the equation. This maneuver groups the logarithmic expressions together, paving the way for simplification. The equation then transforms into $\log _2 x - \log _2(3x + 5) = 4$. The next key step involves using the quotient rule of logarithms. This rule is a cornerstone in manipulating logarithmic expressions, stating that the logarithm of a quotient is equal to the difference of the logarithms. Mathematically, this is expressed as $\log_b a - \log_b c = \log_b(\frac{a}{c})$. Applying this rule to Elsa's equation, we can condense the left side into a single logarithmic term. This transformation yields $\log _2(\frac{x}{3x+5}) = 4$. By applying the quotient rule, we've effectively simplified the equation, making it more manageable. This is a common and powerful technique when dealing with logarithmic equations.

To transform the logarithmic equation into a more solvable form, we need to eliminate the logarithm altogether. The definition of a logarithm provides the perfect tool for this purpose. Remember, the logarithm $\log_b a = c$ is equivalent to the exponential form $b^c = a$. In simpler terms, the logarithm answers the question, "To what power must we raise the base ($b$) to get $a$?" Applying this understanding to Elsa's equation, $\log _2(\frac{x}{3x+5}) = 4$, we recognize that the base is 2, and the exponent is 4. Thus, we can rewrite the equation in exponential form as $2^4 = \frac{x}{3x+5}$. This transformation is a crucial step because it removes the logarithm, turning the equation into a rational equation, which we can solve using standard algebraic techniques. Exponential forms and logarithmic forms are two sides of the same coin, and the ability to move fluently between them is vital for solving logarithmic equations. Now, we have a clearer path forward. The equation $2^4 = \frac{x}{3x+5}$ simplifies to $16 = \frac{x}{3x+5}$. This form is much easier to manipulate algebraically.

To solve this rational equation, our next step is to eliminate the fraction. We achieve this by multiplying both sides of the equation by the denominator, which is $(3x + 5)$. This multiplication gives us $16(3x + 5) = x$. Now, the equation looks much friendlier; we have a linear equation that we can solve by expanding, simplifying, and isolating $x$. When multiplying both sides by $(3x+5)$, we're essentially undoing the division, which allows us to work with a more straightforward algebraic structure. This is a standard method for tackling rational equations, and it's rooted in the basic principles of algebraic manipulation. Distributing the 16 across the terms inside the parenthesis, we obtain $48x + 80 = x$. This equation now involves only linear terms, making it significantly easier to solve for $x$. The transformation from the logarithmic equation to this linear equation is a testament to the power of logarithmic and algebraic techniques. We have successfully moved from a complex logarithmic problem to a familiar linear equation, paving the way for a solution.

Now that we have transformed the equation, $48x + 80 = x$, the next step is to isolate $x$ and find its value. This process involves standard algebraic manipulations. First, subtract $x$ from both sides to group the $x$ terms together, resulting in $47x + 80 = 0$. Next, subtract 80 from both sides to isolate the term with $x$, giving us $47x = -80$. Finally, divide both sides by 47 to solve for $x$, which yields $x = \frac{-80}{47}$. This value seems like a solution, but we're not done yet. In the realm of logarithmic equations, it's crucial to check for extraneous solutions. Extraneous solutions are values that emerge from the algebraic process but do not satisfy the original equation's domain restrictions. Logarithmic functions have a strict requirement: the argument (the value inside the logarithm) must be positive. If a solution makes any argument in the original equation non-positive, it is an extraneous solution and must be discarded.

To check whether $x = \frac{-80}{47}$ is a valid solution, we substitute this value back into the original equation, $\log _2 x=\log _2(3 x+5)+4$, and evaluate the arguments of the logarithms. First, consider $\log _2 x$. Substituting $x = \frac{-80}{47}$, we get $\log _2(\frac{-80}{47})$. Since $\frac{-80}{47}$ is a negative number, and logarithms are undefined for negative numbers, this immediately tells us that $x = \frac{-80}{47}$ is not a valid solution. There's no need to check the other logarithmic term because, as soon as one logarithm is undefined, the entire equation is undefined. This is a key point to remember: a single undefined logarithmic term renders the whole equation meaningless. The process of checking for extraneous solutions is vital because it ensures that our algebraic solution is also a solution in the context of the original logarithmic equation. By verifying the solution against the domain restrictions, we avoid falling into the trap of accepting false answers.

In conclusion, Elsa's assessment that the equation $\log _2 x=\log _2(3 x+5)+4$ has no solution is indeed correct. Our rigorous examination, involving logarithmic transformations, algebraic manipulations, and a crucial check for extraneous solutions, has led us to this definitive answer. We began by applying the quotient rule of logarithms to simplify the equation, then converted it into exponential form to eliminate the logarithms. Solving the resulting rational equation, we arrived at a potential solution, $x = \frac{-80}{47}$. However, the critical step of checking for extraneous solutions revealed that this value is not within the domain of the logarithmic functions in the original equation. Specifically, substituting $x = \frac{-80}{47}$ into $\log _2 x$ resulted in taking the logarithm of a negative number, which is undefined.

This exploration underscores the importance of not only mastering algebraic techniques but also understanding the fundamental properties and restrictions of mathematical functions, such as logarithms. Solving logarithmic equations requires a blend of algebraic skill and a keen awareness of domain limitations. The check for extraneous solutions is not merely a formality; it's an essential safeguard against accepting invalid answers. Elsa's initial claim highlights a common challenge in mathematics: the need to verify solutions within the original context of the problem. This principle applies broadly across various mathematical domains and is a hallmark of rigorous mathematical thinking. The process of solving this equation has been more than just finding an answer; it's been an exercise in mathematical reasoning and precision. The journey from the initial equation to the final conclusion has reinforced the significance of each step in the problem-solving process. Elsa's equation, while seemingly simple, has offered valuable insights into the nature of logarithmic functions and the importance of careful analysis in mathematics.

Therefore, the final answer is that there is no solution to the equation.