Solving Logarithmic Inequalities Graphically

Dec 18, 2025 by Andrew McMorgan 45 views

Hey math whizzes! Ever stared at a logarithmic inequality and thought, "How on earth do I even start solving this?" You're not alone, guys. Sometimes, these problems can look like a tangled mess of numbers and functions. But what if I told you there's a super neat way to tackle them using graphs? Yep, we're talking about graphing inequalities to find those elusive solution sets. Today, we're diving deep into how to solve an inequality like $\log (x+90) \geq 7$ by turning it into a visual quest. Forget the confusing algebra for a sec, and let's explore the power of visualization. It’s all about transforming the abstract into something you can actually see.

So, the big question is: Which pair of related inequalities can be graphed to find the solution set to $\log (x+90) \geq 7$? This isn't just about finding the answer; it's about understanding the method. When we graph, we're essentially looking for the region where the conditions of our inequality are met. For a single inequality like $\log (x+90) \geq 7$, we can rewrite it by introducing a "dummy" variable, usually 'y'. This allows us to break down the original inequality into two separate inequalities involving 'y' and the logarithmic expression. Think of it like this: we're creating a system of inequalities that, when graphed, will show us the solution.

Let's break down the original inequality: $\log (x+90) \geq 7$. The core of this is the comparison between the logarithmic function, $\log (x+90)$, and the constant value, 7. To graph this, we typically set up a system where 'y' represents one side of the comparison, and the other side represents the boundary. So, we're looking for points (x, y) that satisfy both conditions. The inequality $\log (x+90) \geq 7$ means we are interested in the values of 'x' for which the output of the logarithmic function is greater than or equal to 7. When we introduce 'y', we can think of the logarithmic function as $\mathbf{y = \log (x+90)}$. Now, the original inequality becomes $\mathbf{y \geq 7}$. Therefore, the solution set to the original inequality corresponds to the x-values where the graph of $\mathbf{y = \log (x+90)}$ lies on or above the horizontal line $\mathbf{y = 7}$.

This is where the choices come in. We need to pick the pair of inequalities that accurately represent this graphical approach. Let's analyze the options:

Option A: $y \geq \log (x+90)$ and $y \leq 7$ This option suggests we're looking for the region where 'y' is above the curve $\log (x+90)$ AND below the line y=7. This would be the intersection of these two regions. This doesn't match our goal of finding where $\log (x+90)$ is greater than or equal to 7.
Option B: $y \geq \log (x+90)$ and $y \geq 7$ Here, we're considering the region where 'y' is above the curve $\log (x+90)$ AND above the line y=7. This also doesn't align with our original inequality, which requires the logarithmic expression itself to be greater than or equal to 7.
Option C: $y \leq \log (x+90)$ and $y \leq 7$ This option looks for the region where 'y' is below the curve $\log (x+90)$ AND below the line y=7. Again, this is not what we want.
Option D: $y \leq \log (x+90)$ and $y \geq 7$ This is the crucial one, guys. Let's think about what this pair means. We have $\mathbf{y \leq \log (x+90)}$. This inequality describes the region on and below the curve of the logarithmic function $\mathbf{y = \log (x+90)}$. The second inequality is $\mathbf{y \geq 7}$, which describes the region on and above the horizontal line $\mathbf{y = 7}$.

Now, how does this relate back to our original problem, $\log (x+90) \geq 7$? Remember, we introduced 'y' to help us graph. The original inequality is fundamentally asking: 'For what x-values is the value of the function $\log (x+90)$ greater than or equal to 7?'

When we graph the function $\mathbf{y = \log (x+90)}$ and the line $\mathbf{y = 7}$, the solution to $\log (x+90) \geq 7$ lies in the region where the graph of the function is at or above the line y=7. This means the 'y' values of the function are greater than or equal to 7.

Let's re-evaluate how the system of inequalities represents this. If we are looking for the solution to $\log (x+90) \geq 7$, we can set $y = \log (x+90)$. Then the inequality becomes $y \geq 7$. The graph of $y = \log (x+90)$ is the curve we're interested in. The inequality $y \geq 7$ represents the region above or on the horizontal line y=7. The solution to the original inequality corresponds to the x-values where the graph of $y = \log (x+90)$ is on or above the line $y = 7$.

So, when we graph two inequalities to find the solution to one, we are essentially setting up a visual comparison. The goal is to find the 'x' values where the function's output is greater than or equal to 7. This happens when the curve $y = \log (x+90)$ is at or above the line $y = 7$.

Consider the intersection of the regions defined by the inequalities in Option D: $y \leq \log (x+90)$ and $y \geq 7$. The region $y \geq 7$ is above the line y=7. The region $y \leq \log (x+90)$ is below the curve $y = \log (x+90)$. The intersection of these two regions is where the curve $y = \log (x+90)$ is above the line $y = 7$. This is almost what we want, but the inequality signs are flipped compared to how we initially set up y.

Let's revisit the core idea. To solve $\log (x+90) \geq 7$ graphically, we want to find the x-values where the graph of $y = \log (x+90)$ is at or above the horizontal line $y = 7$.

This means we are looking for the region where the y-coordinate of the function is greater than or equal to 7. So, we have:

The curve: $y = \log (x+90)$
The boundary line: $y = 7$

We are interested in the x-values where the curve's y-values are $\mathbf{\geq 7}$. This implies that the solution lies in the region where $y \geq \log (x+90)$ and $y \geq 7$. This still feels a bit off based on the typical way these problems are presented.

Let's rethink the setup for graphical solutions of inequalities. When we solve an inequality like $f(x) \geq c$, we often graph $y = f(x)$ and $y = c$. The solution is where the graph of $y = f(x)$ is on or above the line $y = c$.

In our case, $f(x) = \log (x+90)$ and $c = 7$. So we graph $y = \log (x+90)$ and $y = 7$. We need the region where $y = \log (x+90)$ is on or above $y = 7$. This means we are interested in the points (x, y) such that $y \geq 7$ AND $y = \log (x+90)$. However, the question asks for a pair of inequalities that can be graphed. This suggests we're looking for a region.

Let's consider the standard approach when converting an inequality like $A \geq B$ into a system for graphing. We can introduce 'y' in a way that represents the relationship. Often, we might consider the region bounded by the functions. If we are looking for where $\log (x+90) \geq 7$, this means the output of the logarithm function must be greater than or equal to 7. This is a condition on the 'y' value associated with the function $y = \log (x+90)$.

So, we are interested in the points on the curve $y = \log (x+90)$ where $y \geq 7$. The question implies we should graph two inequalities. This usually means we are looking for the intersection of two shaded regions.

If we set $y = \log (x+90)$, the original inequality becomes $y \geq 7$. We are looking for the x-values where the curve $y = \log (x+90)$ lies on or above the line $y = 7$.

Let's re-examine Option D: $y \leq \log (x+90)$ and $y \geq 7$.

The region $y \geq 7$ is the area above the horizontal line $y = 7$. The region $y \leq \log (x+90)$ is the area below the curve $y = \log (x+90)$.

The intersection of these two regions represents the points (x, y) where $7 \leq y \leq \log (x+90)$. This means the curve $y = \log (x+90)$ must be above or equal to the line $y = 7$. This is precisely where the condition $\log (x+90) \geq 7$ is satisfied.

Think about it this way: We are trying to find the x-values for which the y-value of the function $y = \log (x+90)$ is greater than or equal to 7. When we graph the function $y = \log (x+90)$ and the line $y = 7$, the solution occurs where the curve is above or on the line. The inequalities in option D define a region. The top boundary of this region is the curve $y = \log (x+90)$, and the bottom boundary is the line $y = 7$. The region BETWEEN these boundaries (inclusive) is where $7 \leq y \leq \log (x+90)$. This condition, when we consider the points that lie on the curve $y = \log (x+90)$, directly translates to $\log (x+90) \geq 7$.

So, when graphing the pair of inequalities $y \leq \log (x+90)$ and $y \geq 7$, the area of intersection of these two shaded regions will contain the points (x, y) that satisfy both conditions. Crucially, the x-values within this intersection region are the solutions to the original inequality $\log (x+90) \geq 7$. It's a slightly indirect way of setting it up, but it's a common method for visualizing solutions to inequalities by defining a bounded region.

Let's confirm this with a concrete example. Suppose we wanted to solve $x^2 \geq 4$. We could graph $y = x^2$ and $y = 4$. The solution is where the parabola is on or above the line y=4. This corresponds to the x-values where $|x| \geq 2$. If we were to use a system of inequalities, we might consider $y \geq x^2$ and $y \geq 4$. The intersection would be the region above both. Or, we could consider $y \leq x^2$ and $y \geq 4$. The intersection would be the region between y=4 and $y = x^2$, which is exactly where $x^2 \geq 4$. This confirms the logic of Option D.

Therefore, the correct pair of related inequalities that can be graphed to find the solution set to $\log (x+90) \geq 7$ is $y \leq \log (x+90)$ and $y \geq 7$. By shading the region above $y = 7$ and below $y = \log (x+90)$, the x-values corresponding to this intersection are our solution. It's a clever visual trick that helps solidify your understanding of inequality solutions. Keep practicing this, and you'll be a graphing pro in no time, guys!