Domain Of Logarithmic Functions: F(x) = Log(-9x)

Let's dive into the fascinating world of logarithmic functions and, more specifically, how to pinpoint the domain of a logarithmic function. Our focus today is on a particular example: $f(x) = \log (-9x)$. Understanding the domain is absolutely crucial because it tells us all the possible input values (the 'x' values) for which the function is defined and produces a real number output. For logarithms, there's a fundamental rule: the argument of the logarithm (the part inside the parentheses) must be strictly greater than zero. This is because logarithms are essentially the inverse of exponentiation, and you can't raise a base to any real power and get a non-positive result (except in specific complex number scenarios, which we won't delve into here). So, for $f(x) = \log (-9x)$, the expression $-9x$ is our argument. To find the domain, we need to ensure that $-9x > 0$. This inequality is the key to unlocking the valid 'x' values. We'll solve this simple inequality, and the solution set will be our domain.

Now, let's meticulously solve the inequality $-9x > 0$ to determine the domain of the logarithmic function $f(x) = \log (-9x)$. When dealing with inequalities, we must be careful when multiplying or dividing by a negative number. In this case, to isolate 'x', we need to divide both sides of the inequality by $-9$. Remember the golden rule: when you multiply or divide an inequality by a negative number, you must reverse the direction of the inequality sign. So, dividing $-9x > 0$ by $-9$ gives us $x < \frac{0}{-9}$. Simplifying the right side, we get $x < 0$. This means that any 'x' value that is less than zero will make the argument $-9x$ positive, and therefore, the function $f(x) = \log (-9x)$ will be defined. We can express this domain in interval notation as $(-\infty, 0)$. Any number in this interval, when plugged into the function, will yield a real number. For instance, if we pick $x = -1$, the argument becomes $-9(-1) = 9$, and $\log(9)$ is a perfectly valid real number. However, if we were to pick $x = 1$, the argument would be $-9(1) = -9$, and $\log(-9)$ is undefined in the realm of real numbers. This exercise highlights the importance of the argument always being positive.

To further solidify our understanding of the domain of a logarithmic function like $f(x) = \log (-9x)$, let's consider the graphical interpretation. The graph of a logarithmic function, $y = \log_b(x)$, has a vertical asymptote at $x=0$. The domain is restricted to $x > 0$ for a standard $\log(x)$ function. However, in our case, the argument is $-9x$. When the argument of a logarithm is an expression involving 'x', the domain is determined by setting that expression to be greater than zero. For $f(x) = \log (-9x)$, we established that $-9x > 0$, which led us to $x < 0$. This transformation (multiplying 'x' by a negative constant) effectively reflects the standard logarithmic graph across the y-axis. The original $\log(x)$ graph exists only for positive x-values. By introducing $-9x$ as the argument, we are essentially asking: for what 'x' values is $-9x$ positive? This occurs when 'x' itself is negative. The vertical asymptote for this transformed function is still at the value that makes the argument zero, which is $-9x = 0$, so $x=0$. But the function's domain is where the argument is positive, i.e., $x < 0$. So, the graph of $f(x) = \log (-9x)$ exists entirely in the second and third quadrants, mirroring the standard logarithmic curve but on the left side of the y-axis. This visual understanding reinforces the mathematical derivation of the domain.

It's essential to remember that the base of the logarithm also plays a role in its properties, though it doesn't directly affect the domain calculation, which is solely dependent on the argument being positive. For common logarithms (base 10, often written as $\log$), natural logarithms (base $e$, written as $\ln$), or any other base $b > 0$ and $b \neq 1$, the rule remains the same: the argument must be positive. So, whether you see $\log(-9x)$, $\ln(-9x)$, or $\log_{10}(-9x)$, the process for finding the domain is identical. You set the argument, $-9x$, greater than zero and solve for 'x'. The resulting inequality, $x < 0$, defines the set of all valid inputs for our function. This consistency across different bases is a fundamental property that simplifies working with logarithmic functions. Always keep your eyes on that argument – it's the gateway to determining the domain.

Let's consider some additional examples to reinforce the concept of finding the domain of a logarithmic function. Suppose we had $g(x) = \log (x - 3)$. Here, the argument is $(x - 3)$. To find the domain, we set $x - 3 > 0$. Adding 3 to both sides gives us $x > 3$. In interval notation, this is $(3, \infty)$. Now, what about $h(x) = \ln (2x + 5)$? The argument is $(2x + 5)$. We set $2x + 5 > 0$. Subtracting 5 gives $2x > -5$. Dividing by 2, we get $x > -5/2$. The domain is $(-5/2, \infty)$. Finally, let's look at a slightly more complex argument, $k(x) = \log (x^2 - 4)$. Here, the argument is $(x^2 - 4)$. We need $x^2 - 4 > 0$. This inequality can be factored as $(x - 2)(x + 2) > 0$. This inequality holds true when both factors are positive ($x > 2$ and $x > -2$, so $x > 2$) or when both factors are negative ($x < 2$ and $x < -2$, so $x < -2$). Therefore, the domain for $k(x)$ is $(-\infty, -2) \cup (2, \infty)$. These examples illustrate that while the principle remains constant (argument > 0), the complexity of solving the resulting inequality can vary, but the fundamental approach to finding the domain of a logarithmic function is consistent.

In conclusion, determining the domain of a logarithmic function is a straightforward process rooted in a single, crucial rule: the argument of the logarithm must always be strictly greater than zero. For our specific function, $f(x) = \log (-9x)$, this translates to the inequality $-9x > 0$. Solving this inequality, keeping in mind the rule about multiplying or dividing by a negative number, yields $x < 0$. This means that any real number less than zero is a valid input for this function, and all numbers greater than or equal to zero are not. The domain, expressed in interval notation, is $(-\infty, 0)$. This understanding is fundamental for analyzing and graphing logarithmic functions, as it defines the boundaries of where the function is meaningful in the real number system. Always remember to isolate the argument and set it to be greater than zero; this principle will guide you through any logarithmic domain problem.

For further exploration into the properties of logarithms and functions, you can consult resources like Khan Academy, which offers comprehensive explanations and practice exercises on various mathematical topics.