Logarithmic Equation: Passing Through (2,0) & (-1,3)

Nov 14, 2025 by Andrew McMorgan 53 views

Hey Plastik Magazine readers! Ever found yourself scratching your head over logarithmic functions and their transformations? Well, you're in the right place. Today, we're diving deep into how to find the equation of a transformed logarithm that gracefully passes through two specific points: (2,0) and (-1,3). Trust me, it's not as daunting as it sounds. Let's break it down, step by step, in a way that even your grandma could understand. Get ready to unleash your inner mathlete!

Understanding the General Form

Before we jump into the specifics, let's get cozy with the general form of a transformed logarithmic function. This is your trusty map that will guide us through the mathematical wilderness. The general form looks like this:

f(x) = a * log_b(x - h) + k

Where:

f(x) is the value of the function at x.
a is the vertical stretch or compression factor. Think of it as how much the logarithm is squished or stretched vertically.
b is the base of the logarithm. This determines how quickly the function grows or decays.
h is the horizontal shift. It tells us how much the graph has moved left or right along the x-axis.
k is the vertical shift. This indicates how much the graph has moved up or down along the y-axis.

Now that we know what each component represents, let's put this knowledge to practical use. Our mission, should we choose to accept it, is to find the values of a, b, h, and k that make our logarithmic function pass through the points (2,0) and (-1,3). Sounds like fun, right? Let's get started!

Plugging in the Points

Alright, let's get our hands dirty with some good ol' substitution. We know our transformed logarithm passes through the points (2,0) and (-1,3). This means when x = 2, f(x) = 0, and when x = -1, f(x) = 3. Let's plug these values into our general equation. This will give us two equations with four unknowns – a, b, h, and k. Don't worry; we'll tackle this systematically.

For the point (2,0):

0 = a * log_b(2 - h) + k

For the point (-1,3):

3 = a * log_b(-1 - h) + k

Now we have a system of two equations. To solve this, we need to make some strategic moves. The goal is to reduce the number of unknowns and make the equations easier to handle. One common approach is to eliminate one of the variables. In this case, we can eliminate k by subtracting one equation from the other. This will give us a new equation with only a, b, and h.

Eliminating 'k'

Subtract the first equation from the second:

3 - 0 = [a * log_b(-1 - h) + k] - [a * log_b(2 - h) + k]

Simplify:

3 = a * log_b(-1 - h) - a * log_b(2 - h)

Factor out a:

3 = a * [log_b(-1 - h) - log_b(2 - h)]

Now we can use the logarithm property that states log_b(x) - log_b(y) = log_b(x/y):

3 = a * log_b((-1 - h) / (2 - h))

This equation is a bit cleaner, but we still have three unknowns. To proceed, we need to make an assumption or find another piece of information. Often, in these types of problems, the base b is given or can be assumed to be 10 (common logarithm) or e (natural logarithm). Let's assume b = 10 for now. If we have additional information about the base, we can adjust our calculations accordingly.

Assuming b = 10

With b = 10, our equation becomes:

3 = a * log_10((-1 - h) / (2 - h))

Now, let's go back to one of our original equations to express k in terms of a and h. Using the first equation:

0 = a * log_10(2 - h) + k

Solve for k:

k = -a * log_10(2 - h)

Now we have k expressed in terms of a and h. We can substitute this back into our second original equation to eliminate k:

3 = a * log_10(-1 - h) - a * log_10(2 - h)

Which we already simplified to:

3 = a * log_10((-1 - h) / (2 - h))

At this point, we have one equation with two unknowns (a and h). This means we need another equation or a clever trick to solve for both variables. Unfortunately, without additional information, it's difficult to find unique values for a and h. However, we can express a in terms of h or vice versa.

Expressing 'a' in terms of 'h'

From our equation 3 = a * log_10((-1 - h) / (2 - h)), we can solve for a:

a = 3 / log_10((-1 - h) / (2 - h))

Now we have a expressed as a function of h. We can substitute this expression for a back into our equation for k:

k = -[3 / log_10((-1 - h) / (2 - h))] * log_10(2 - h)

So, we have:

a = 3 / log_10((-1 - h) / (2 - h)) k = -[3 / log_10((-1 - h) / (2 - h))] * log_10(2 - h)

Putting It All Together

Our transformed logarithmic function is:

f(x) = [3 / log_10((-1 - h) / (2 - h))] * log_10(x - h) - [3 / log_10((-1 - h) / (2 - h))] * log_10(2 - h)

This equation represents a family of logarithmic functions that pass through the points (2,0) and (-1,3). The specific function depends on the value of h. Without additional information, we cannot determine a unique value for h. However, if we were given another point or a specific transformation (e.g., a vertical stretch factor), we could solve for h and find the unique equation.

For example, if we knew h, we could plug it in and get specific values for a and k. If we assume a value for h, let's say h = 0:

a = 3 / log_10((-1 - 0) / (2 - 0)) = 3 / log_10(-1/2)

Since the logarithm of a negative number is undefined in the real number system, h = 0 is not a valid solution. We need to ensure that (-1 - h) / (2 - h) is positive.

Ensuring a Valid Logarithm

To ensure that the argument of the logarithm is positive, we need to satisfy the following inequality:

(-1 - h) / (2 - h) > 0

This inequality holds when both the numerator and the denominator are either positive or negative. Let's analyze each case:

Case 1: Both positive

-1 - h > 0 => h < -1 2 - h > 0 => h < 2

So, for both to be positive, h < -1.

Case 2: Both negative

-1 - h < 0 => h > -1 2 - h < 0 => h > 2

So, for both to be negative, h > 2.

Therefore, the valid range for h is h < -1 or h > 2. This ensures that the argument of the logarithm is positive, and the function is defined.

Example with h = 3

Let's pick a value for h that satisfies our condition, say h = 3 (since h > 2).

a = 3 / log_10((-1 - 3) / (2 - 3)) = 3 / log_10((-4) / (-1)) = 3 / log_10(4)

k = -a * log_10(2 - h) = -(3 / log_10(4)) * log_10(2 - 3) = -(3 / log_10(4)) * log_10(-1)

Since log_10(-1) is undefined, h = 3 is not a valid solution either. We made an error by not considering the domain of the original equation, namely x - h > 0, which means x > h.

If x = 2, then 2 > h and if x = -1, then -1 > h. Thus, the only valid condition is h < -1.

Let's try h = -2: a = 3 / log_10((-1 - (-2)) / (2 - (-2))) = 3 / log_10(1/4) k = -a * log_10(2 - (-2)) = - (3 / log_10(1/4)) * log_10(4) = - (3 / log_10(1/4)) * log_10(4) = - (3 / log_10(4^{-1})) * log_10(4) = - (3 / (-log_10(4))) * log_10(4) = 3

So, with h = -2, we have a = 3 / log_10(1/4) and k = 3.

Our transformed logarithmic function is:

f(x) = (3 / log_10(1/4)) * log_10(x + 2) + 3

Conclusion

Finding the equation of a transformed logarithm that passes through specific points involves understanding the general form of the logarithmic function, substituting the given points, and solving for the unknown parameters. In many cases, additional information or assumptions are needed to find a unique solution. Remember, the key is to break down the problem into smaller, manageable steps and use the properties of logarithms to simplify the equations. Keep practicing, and you'll become a logarithm master in no time!

And that's a wrap, folks! Hope this deep dive into logarithmic transformations helped clear things up. Keep playing with those equations, and remember, math can be fun! Stay tuned for more mind-bending math adventures in Plastik Magazine!