Inverse Logarithmic Function: Find Values For The Inverse

Hey math lovers! Ever wondered about the inverse of a logarithmic function? Today, we're diving deep into a specific example: $f(x)=\log _{0.5^x}$. We'll uncover its inverse, $f^{-1}(x)=0.5^x$, and then tackle a super fun challenge – completing a table for this inverse function. Get ready to flex those math muscles, guys! Understanding inverse functions is crucial in mathematics, especially when you're dealing with exponential and logarithmic relationships. These pairs of functions essentially 'undo' each other, which makes them incredibly powerful tools for solving equations and analyzing data.

Let's start by breaking down the function we're working with. We have $f(x)=\log _{0.5^x}$. This notation might look a little different from what you're used to. Typically, the base of the logarithm is a constant, like $\log_2 x$ or $\log_{10} x$. Here, the base itself seems to involve $x$. However, there's a property of logarithms that can simplify this: $\log_{b^k} m = \frac{1}{k} \log_b m$. In our case, the base is $0.5^x$. This means we can rewrite the function as $f(x) = \log_{(0.5)^x} x$. Using the property mentioned above, with $b=0.5$ and $k=x$, we get $f(x) = \frac{1}{x} \log_{0.5} x$. This form might be more familiar to some of you. Now, let's talk about finding the inverse. To find the inverse of any function $f(x)$, we follow a standard procedure: replace $f(x)$ with $y$, swap $x$ and $y$, and then solve for $y$.

So, let's apply this to $f(x)=\frac{1}{x} \log_{0.5} x$. First, we set $y = \frac{1}{x} \log_{0.5} x$. Then, we swap $x$ and $y$ to get $x = \frac{1}{y} \log_{0.5} y$. Solving this equation for $y$ is where things can get a bit tricky. It's not always straightforward to isolate $y$ when it appears in both the denominator and within the logarithm. This is why the problem statement gives us a direct hint: the inverse function is $f^{-1}(x)=0.5^x$. Let's quickly verify this. If $f^{-1}(x) = 0.5^x$, then its inverse should be the original function $f(x)$. Let $g(x) = 0.5^x$. To find its inverse, we set $y = 0.5^x$. Swapping $x$ and $y$ gives $x = 0.5^y$. To solve for $y$, we can take the logarithm of both sides. Using the logarithm with base 0.5 is convenient here: $\log_{0.5} x = \log_{0.5} (0.5^y)$. This simplifies to $\log_{0.5} x = y$. So, the inverse of $g(x)=0.5^x$ is $g^{-1}(x) = \log_{0.5} x$.

Now, let's reconcile this with our original function $f(x)=\log _{0.5^x}$. Remember our simplification? $f(x) = \frac{1}{x} \log_{0.5} x$. This means the inverse we found, $g^{-1}(x) = \log_{0.5} x$, is not exactly $f(x)$. There seems to be a misunderstanding in the initial problem statement or the provided inverse. Let's re-examine $f(x)=\log _{0.5^x}$. If we interpret the base as $0.5$ raised to the power of $x$, as in $\log_{(0.5^x)}$, then the simplification $f(x) = \frac{1}{x} \log_{0.5} x$ is correct. However, if the notation implies $f(x) = \log_{0.5} (x)$, which is a more standard form, then its inverse is indeed $f^{-1}(x) = 0.5^x$. Given that the problem explicitly states the inverse is $f^{-1}(x)=0.5^x$, it's highly probable that the intended function was $f(x) = \log_{0.5} x$. Let's proceed with this assumption, as it aligns with the provided inverse.

Understanding the Inverse Function $f^{-1}(x) = 0.5^x$

Alright guys, let's focus on the inverse function we've been given: $f^{-1}(x)=0.5^x$. This is an exponential function. Remember, exponential functions are the direct counterparts to logarithmic functions. When you graph $y = b^x$, you see a curve that either increases rapidly (if $b > 1$) or decreases rapidly (if $0 < b < 1$). In our case, the base is $0.5$, which is between 0 and 1. This means the graph of $f^{-1}(x)=0.5^x$ will be a decreasing curve. It will pass through the point $(0, 1)$ because any non-zero number raised to the power of 0 is 1. As $x$ gets larger in the positive direction, $0.5^x$ gets closer and closer to 0. As $x$ gets larger in the negative direction (i.e., more negative), $0.5^x$ grows larger and larger.

Now, the key property of inverse functions is that if $(x, y)$ is a point on the graph of $f(x)$, then $(y, x)$ is a point on the graph of $f^{-1}(x)$. Conversely, if $(a, b)$ is a point on the graph of $f^{-1}(x)$, then $(b, a)$ is a point on the graph of $f(x)$. Our task is to complete a table for the inverse function $f^{-1}(x)=0.5^x$. The table provides some $x$ values and corresponding $y$ values for this inverse function. We need to find the missing values. The table looks like this:

\begin{tabular}{|c|c|c|c|c|c|} \hline $x$ & -2 & -1 & 0 & 1 & 2 \ \hline $y$ & 4 & $-a$ & $b$ & $c$ & 0.25 \ \hline \end{tabular}

We are given the function $f^{-1}(x) = 0.5^x$. We need to plug in the given $x$ values into this function to find the corresponding $y$ values. Let's go through it step-by-step, keeping in mind that $0.5$ can also be written as $\frac{1}{2}$.

Calculating Table Values for $f^{-1}(x) = 0.5^x$

Let's calculate the $y$ values for each given $x$ value using $f^{-1}(x) = 0.5^x$:

• When $x = -2$: $y = f^{-1}(-2) = 0.5^{-2}$. Remember that a negative exponent means we take the reciprocal of the base raised to the positive exponent. So, $0.5^{-2} = (\frac{1}{2})^{-2}$. This is equal to $(\frac{2}{1})^2 = 2^2 = 4$. So, when $x=-2$, $y=4$. This matches the first entry in our table, which is awesome! It confirms we're on the right track. This pair $(-2, 4)$ is a point on the graph of $f^{-1}(x)$. Consequently, the point $(4, -2)$ must be on the graph of the original function $f(x) = \log_{0.5} x$. This relationship is the heart of understanding inverse functions – how the coordinates swap.

• When $x = -1$: $y = f^{-1}(-1) = 0.5^{-1}$. Using the same logic as before, $0.5^{-1} = (\frac{1}{2})^{-1}$. This equals $(\frac{2}{1})^1 = 2^1 = 2$. So, when $x=-1$, $y=2$. Looking at our table, the value is given as $-a$. This means $-a = 2$. To find $a$, we multiply both sides by -1: $a = -2$. So, the value in the table is $y = -(-2) = 2$. The entry in the table is actually $-a$, so we are solving for $a$. The $y$ value is 2. Therefore, $-a=2$, which means $a=-2$. This is a crucial step where we find the value of $a$. The point $(-1, 2)$ is on the graph of $f^{-1}(x)$. The corresponding point on $f(x)$ would be $(2, -1)$. Isn't it cool how the signs work out?

• When $x = 0$: $y = f^{-1}(0) = 0.5^0$. Any non-zero number raised to the power of 0 is 1. So, $y = 1$. The table shows this value as $b$. Therefore, $b = 1$. The point $(0, 1)$ is on the graph of $f^{-1}(x)$. This means the point $(1, 0)$ is on the graph of $f(x) = \log_{0.5} x$, which is correct since $\log_{0.5} 1 = 0$. This point is the y-intercept for the exponential function and the x-intercept for the logarithmic function.

• When $x = 1$: $y = f^{-1}(1) = 0.5^1$. Any number raised to the power of 1 is itself. So, $y = 0.5$. The table shows this value as $c$. Therefore, $c = 0.5$. The point $(1, 0.5)$ is on the graph of $f^{-1}(x)$. This implies that the point $(0.5, 1)$ is on the graph of $f(x) = \log_{0.5} x$. Let's check: $\log_{0.5} 0.5 = 1$, which is true. So, our value for $c$ is correct.

• When $x = 2$: $y = f^{-1}(2) = 0.5^2$. This means $0.5 \times 0.5 = 0.25$. So, when $x=2$, $y=0.25$. This matches the last entry in our table. The point $(2, 0.25)$ is on the graph of $f^{-1}(x)$. This means the point $(0.25, 2)$ is on the graph of $f(x) = \log_{0.5} x$.

Completing the Table

So, let's put it all together! We've calculated the values needed to complete the table for the inverse function $f^{-1}(x)=0.5^x$.

First, we found that for $x=-1$, the $y$ value is $2$. Since the table shows $-a$, we have $-a = 2$, which means $a = -2$.

Next, for $x=0$, the $y$ value is $1$. The table shows this as $b$, so $b = 1$.

Finally, for $x=1$, the $y$ value is $0.5$. The table shows this as $c$, so $c = 0.5$.

Here's the completed table:

\begin{tabular}{|c|c|c|c|c|c|} \hline $x$ & -2 & -1 & 0 & 1 & 2 \ \hline $y$ & 4 & 2 & 1 & 0.5 & 0.25 \ \hline \end{tabular}

And if we specifically fill in the variables $a$, $b$, and $c$ as requested:

$a = -2$

$b = 1$

$c = 0.5$

This exercise really highlights how inverse functions work. For every point $(x, y)$ on the graph of $f^{-1}(x) = 0.5^x$, the point $(y, x)$ is on the graph of $f(x) = \log_{0.5} x$. For instance, we found the point $(-1, 2)$ is on the graph of $f^{-1}(x)$. This means the point $(2, -1)$ must be on the graph of $f(x)$. Let's check: $f(2) = \log_{0.5} 2$. Since $0.5 = \frac{1}{2}$, we have $\log_{1/2} 2$. What power do we raise $1/2$ to get 2? It's $-1$, because $(\frac{1}{2})^{-1} = 2$. So, $f(2) = -1$, which confirms our point $(2, -1)$ is indeed on the graph of $f(x)$. This reciprocal relationship between the coordinates is the fundamental concept behind inverse functions and is super useful for solving problems and understanding function behavior. Keep practicing, and these concepts will become second nature!