Solving Logarithmic Equations: Exact Solutions

When we talk about solving logarithmic equations, we're diving into a fascinating area of mathematics where we unravel the mysteries hidden within logarithmic functions. The core idea behind solving logarithmic equations is to use the properties of logarithms to isolate the variable, typically 'x', and find its exact value. It's like being a detective, piecing together clues to arrive at the truth. We often encounter equations that look complex, such as $\ln x+\ln (x-10)=\ln (9 x-88)$. To tackle this, we first need to understand the fundamental properties of logarithms. The product rule, $\ln a + \ln b = \ln (ab)$, is a game-changer here. It allows us to combine multiple logarithmic terms on one side of the equation into a single logarithm. So, applying this to our example, the left side, $\ln x+\ln (x-10)$, becomes $\ln [x(x-10)]$. This simplification is crucial because it brings us closer to an equation where we can equate the arguments of the logarithms. Remember, if $\ln A = \ln B$, then it must be true that $A = B$. This is the cornerstone of solving these types of equations. However, we must always be mindful of the domain of logarithmic functions. The argument of a logarithm must always be positive. This means that for $\ln x$ to be defined, $x > 0$. Similarly, for $\ln (x-10)$, we need $x-10 > 0$, which implies $x > 10$. And for $\ln (9x-88)$, we require $9x-88 > 0$, meaning $x > 88/9$. The most restrictive of these conditions is $x > 10$. Any solution we find must satisfy this condition; otherwise, it's an extraneous solution and must be discarded. This pre-calculation of the domain is a vital step that prevents us from falling into mathematical traps.

Applying Logarithmic Properties to Simplify

Let's get back to our specific equation: $\ln x+\ln (x-10)=\ln (9 x-88)$. We've already established that the sum of the logarithms on the left can be combined using the product rule: $\ln [x(x-10)] = \ln (9 x-88)$. Now, the equation is in a much more manageable form. Since the logarithms on both sides have the same base (the natural logarithm, base 'e'), we can equate their arguments. This gives us a polynomial equation: $x(x-10) = 9x-88$. Expanding the left side, we get $x^2 - 10x = 9x-88$. Our next step is to rearrange this into a standard quadratic equation form, $ax^2+bx+c=0$. To do this, we move all terms to one side: $x^2 - 10x - 9x + 88 = 0$. This simplifies to $x^2 - 19x + 88 = 0$. Now, we have a quadratic equation that we can solve using various methods, such as factoring, completing the square, or the quadratic formula. Factoring is often the quickest if the equation is factorable. We need to find two numbers that multiply to 88 and add up to -19. After a bit of thought, we find that -8 and -11 fit the bill: $(-8) \times (-11) = 88$ and $(-8) + (-11) = -19$. Therefore, we can factor the quadratic equation as $(x-8)(x-11) = 0$. Setting each factor to zero, we get two potential solutions: $x-8=0$ which yields $x=8$, and $x-11=0$ which yields $x=11$. This is where the domain we established earlier becomes critically important. We must check if these potential solutions satisfy the condition $x > 10$. Looking at our potential solutions, $x=8$ does not satisfy $x > 10$. If we were to plug $x=8$ back into the original equation, the term $\ln (x-10)$ would become $\ln (8-10) = \ln (-2)$, which is undefined. Thus, $x=8$ is an extraneous solution and must be rejected. On the other hand, $x=11$ does satisfy $x > 10$. Let's quickly verify it in the original equation: $\ln 11 + \ln (11-10) = \ln 11 + \ln 1 = \ln 11$. And on the right side: $\ln (9 \times 11 - 88) = \ln (99 - 88) = \ln 11$. Since both sides are equal, $x=11$ is a valid solution.

Verifying Solutions and Understanding Extraneous Roots

The process of verifying solutions in logarithmic equations is not just a formality; it's an essential step to ensure the validity of our results. As we saw with the equation $\ln x+\ln (x-10)=\ln (9 x-88)$, we arrived at two potential solutions, $x=8$ and $x=11$, by solving the derived quadratic equation. However, the domain of logarithmic functions dictates that their arguments must be strictly positive. For $\ln x$, we need $x > 0$. For $\ln (x-10)$, we need $x-10 > 0$, which means $x > 10$. And for $\ln (9x-88)$, we need $9x-88 > 0$, which implies $x > 88/9 \approx 9.78$. The most stringent condition among these is $x > 10$. This is our domain constraint. Any solution that falls outside this domain is considered an extraneous solution. Extraneous solutions arise when we manipulate equations, especially by squaring both sides or, in this case, by equating arguments of logarithms. When we go from $\ln A = \ln B$ to $A=B$, we are essentially removing the logarithmic function. This process can sometimes introduce solutions that satisfy $A=B$ but not the original logarithmic form because one or both of $A$ or $B$ might be non-positive for those values. In our specific problem, $x=8$ is less than 10. If we substitute $x=8$ into the original equation, we would have $\ln 8 + \ln (8-10) = \ln 9(8)-88$. This simplifies to $\ln 8 + \ln (-2) = \ln (72-88)$, or $\ln 8 + \ln (-2) = \ln (-16)$. Since the logarithm of a negative number is undefined in the real number system, $x=8$ is not a valid solution. It's an extraneous root introduced during the simplification process. Now, let's consider $x=11$. This value satisfies our domain constraint $x > 10$. Substituting $x=11$ into the original equation: $\ln 11 + \ln (11-10) = \ln 9(11)-88$. This becomes $\ln 11 + \ln 1 = \ln (99-88)$. Since $\ln 1 = 0$, the left side is $\ln 11 + 0 = \ln 11$. The right side is $\ln 11$. Because $\ln 11 = \ln 11$, the solution $x=11$ is valid. The solution set for the equation is therefore just {$11$}. It's crucial to remember that every solution obtained from the algebraic manipulation must be checked against the domain of the original logarithmic equation. This verification step ensures that we only include solutions that are mathematically sound and do not lead to undefined terms. The beauty of mathematics lies in its rigor, and checking for extraneous solutions is a prime example of this.

The Importance of Exact Solutions

In mathematics, especially in fields like calculus, physics, and engineering, the emphasis on exact solutions is paramount. When we solve an equation like $\ln x+\ln (x-10)=\ln (9 x-88)$, finding the exact solution means expressing the answer in its most precise form, free from approximations or rounding errors. For this particular problem, the exact solution we found is $x=11$. This is a clean, integer value, and it's already in its exact form. However, in many other logarithmic equations, the solutions might involve irrational numbers, such as square roots or transcendental numbers like $\pi$ or $e$. For example, if a solution turned out to be $\sqrt{2}$ or $3\ln 5$, we would leave it in that form rather than substituting a decimal approximation like $1.414$ or $4.828$. Using approximations can lead to inaccuracies that can propagate through subsequent calculations, potentially leading to incorrect conclusions or flawed designs in practical applications. Imagine calculating the trajectory of a satellite using rounded values for a crucial parameter; the error could be significant. Therefore, whenever possible, we strive to maintain the exact form of the solution. This involves using properties of logarithms and algebraic manipulations judiciously to simplify the equation without losing precision. Techniques like the quadratic formula can yield solutions involving radicals, like $x = \frac{19 \pm \sqrt{19^2 - 4(1)(88)}}{2} = \frac{19 \pm \sqrt{361 - 352}}{2} = \frac{19 \pm \sqrt{9}}{2} = \frac{19 \pm 3}{2}$. This gives us $x = \frac{19+3}{2} = \frac{22}{2} = 11$ and $x = \frac{19-3}{2} = \frac{16}{2} = 8$. Here, the exact solutions are expressed using integers and basic arithmetic operations. If the discriminant (the part under the square root) were not a perfect square, the exact solution would involve that radical, e.g., $x = \frac{19 \pm \sqrt{17}}{2}$. Leaving it in this form ensures maximum accuracy. The solution set for our problem, containing only the valid exact solutions, is {$11$}. This commitment to exactness is a hallmark of rigorous mathematical practice and is essential for reliable scientific and engineering endeavors. It's about preserving the integrity of the mathematical result from its origin to its final application. For more on the properties of logarithms and solving equations, you can visit Khan Academy's Mathematics section.