Logarithmic Equation: Solve $\log_2(y+14)-\log_2(y)=11$

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When we encounter a logarithmic equation like $\log_2(y+14)-\log_2(y)=11$, the first thing that comes to mind is how to isolate the variable 'y' and find its value. This type of problem is common in mathematics, especially in algebra and calculus, and understanding how to solve it involves a few key properties of logarithms. The goal here is to simplify the equation using these properties and then convert it into a more manageable form, typically an exponential equation, which we can then solve for 'y'. So, let's dive into the process of solving this specific logarithmic equation, breaking down each step to make it clear and easy to follow. We'll be using the properties of logarithms to combine the terms and then transform the equation into a form where we can directly solve for 'y'.

Understanding Logarithm Properties for Equation Solving

Before we get our hands dirty with the specific equation $\log_2(y+14)-\log_2(y)=11$, it's crucial to have a solid grasp of the fundamental properties of logarithms. These properties are our toolkit for manipulating logarithmic expressions. One of the most important properties for this particular problem is the quotient rule, which states that $\log_b(M) - \log_b(N) = \log_b(M/N)$. This rule allows us to combine two logarithmic terms with the same base that are being subtracted into a single logarithmic term. In our equation, we have $\log_2(y+14)-\log_2(y)$, which perfectly fits the pattern for applying the quotient rule. By using this rule, we can rewrite the left side of the equation as a single logarithm: \log_2\left(rac{y+14}{y}\right). This simplification is a critical first step because it reduces the complexity of the equation, bringing us closer to isolating 'y'.

Another key property we'll implicitly use is the definition of a logarithm itself. The equation $\log_b(x) = a$ is equivalent to the exponential equation $b^a = x$. This is the bridge that allows us to move from a logarithmic equation to an exponential one, which is often easier to solve. In our case, once we've applied the quotient rule, we'll have an equation in the form $\log_2( ext{expression}) = 11$. Using the definition, we can then rewrite this as $2^{11} = ext{expression}$. This transformation is essential because exponential equations can usually be solved using algebraic techniques.

It's also important to remember the domain of logarithmic functions. The argument of a logarithm must always be positive. In our original equation, we have two logarithmic terms: $\log_2(y+14)$ and $\log_2(y)$. This means that we must satisfy two conditions: $y+14 > 0$ and $y > 0$. Combining these, we find that $y$ must be greater than 0 for both logarithms to be defined. This is a vital check we'll perform at the end to ensure our solution is valid.

So, with these properties in mind – the quotient rule for combining logarithms and the definition for converting to exponential form, along with the domain considerations – we are well-equipped to tackle the equation $\log_2(y+14)-\log_2(y)=11$ systematically.

Step-by-Step Solution Process

Let's begin by applying the quotient rule for logarithms to the left side of the equation $\log_2(y+14)-\log_2(y)=11$. As we discussed, this rule states that $\log_b(M) - \log_b(N) = \log_b(M/N)$. Applying this to our equation, we get:

$\log_2\left(\frac{y+14}{y}\right) = 11$

Now, we have a single logarithmic term. The next step is to convert this logarithmic equation into an equivalent exponential equation. We use the definition of a logarithm: if $\log_b(x) = a$, then $b^a = x$. In our case, the base 'b' is 2, the exponent 'a' is 11, and the argument 'x' is $\frac{y+14}{y}$. So, we can rewrite the equation as:

$2^{11} = \frac{y+14}{y}$

We know that $2^{11}$ is equal to 2048. So the equation becomes:

$2048 = \frac{y+14}{y}$

To solve for 'y', we need to eliminate the denominator. We can do this by multiplying both sides of the equation by 'y'. It's important to note here that we are assuming $y \neq 0$. We will later check if our final solution satisfies the domain requirements for the original logarithmic equation, which implicitly includes $y > 0$.

$2048y = y + 14$

Now, we want to gather all terms containing 'y' on one side of the equation and the constant terms on the other. Subtract 'y' from both sides:

$2048y - y = 14$

$2047y = 14$

Finally, to isolate 'y', divide both sides by 2047:

$y = \frac{14}{2047}$

Verifying the Solution and Domain Considerations

Now that we have found a potential solution, $y = \frac{14}{2047}$, it's absolutely essential to verify it. This involves two main parts: checking if it satisfies the original equation and ensuring it meets the domain requirements for the logarithmic functions involved. Without this verification step, our solution might be extraneous, meaning it works in an intermediate step but not in the original problem.

First, let's consider the domain. The original equation contains $\log_2(y+14)$ and $\log_2(y)$. For these logarithms to be defined, their arguments must be strictly positive. This means we need:

1. $y+14 > 0

2. $y > 0$

Let's test our solution $y = \frac{14}{2047}$:

1. Is $y > 0$? Yes, $\frac{14}{2047}$ is a positive number.

2. Is $y+14 > 0$? Since $y$ is positive, adding 14 to it will definitely result in a positive number. $\frac{14}{2047} + 14 = \frac{14 + 14 \times 2047}{2047} = \frac{14 + 28658}{2047} = \frac{28672}{2047}$, which is positive.

Since our solution $y = \frac{14}{2047}$ satisfies both conditions, it is within the domain of the original equation. This is a crucial check that eliminates potential extraneous solutions that might arise during the solving process.

Next, let's substitute $y = \frac{14}{2047}$ back into the original equation $\log_2(y+14)-\log_2(y)=11$ to see if it holds true.

We already calculated $y+14 = \frac{28672}{2047}$ and $y = \frac{14}{2047}$.

So, the left side of the equation becomes:

$\log_2\left(\frac{28672}{2047}\right) - \log_2\left(\frac{14}{2047}\right)$

Using the quotient rule for logarithms, $\log_b(M) - \log_b(N) = \log_b(M/N)$, we can combine these terms:

$\log_2\left(\frac{\frac{28672}{2047}}{\frac{14}{2047}}\right)$

When dividing fractions, we multiply by the reciprocal of the denominator:

$\log_2\left(\frac{28672}{2047} \times \frac{2047}{14}\right)$

The 2047s cancel out:

$\log_2\left(\frac{28672}{14}\right)$

Now, let's perform the division $28672

\div 14$:

$28672

\div 14 = 2048$

So the expression simplifies to:

$\log_2(2048)$

We need to find what power we must raise 2 to in order to get 2048. We know that $2^{10} = 1024$, and $2^{11} = 1024 \times 2 = 2048$.

Therefore, $\log_2(2048) = 11$.

This matches the right side of our original equation. Thus, our solution $y = \frac{14}{2047}$ is correct and is not an extraneous solution.

Conclusion and Further Exploration

In summary, by carefully applying the properties of logarithms, specifically the quotient rule to combine terms and the definition to convert the logarithmic equation into an exponential one, we successfully solved the equation $\log_2(y+14)-\log_2(y)=11$. The solution we found is $y = \frac{14}{2047}$. We also emphasized the critical importance of verifying the solution against the original equation and ensuring it satisfies the domain requirements for logarithmic functions, which in this case meant $y > 0$. This verification step is crucial for avoiding extraneous solutions that can sometimes arise from algebraic manipulations.

Logarithmic equations are a fundamental part of mathematics, appearing in various fields such as finance (compound interest calculations), science (measuring sound intensity or earthquake magnitudes), and computer science (analyzing algorithm efficiency). Understanding how to manipulate and solve these equations opens doors to comprehending complex phenomena and models.

If you're interested in learning more about logarithms, their properties, and their applications, I highly recommend exploring resources that delve deeper into these topics. For instance, understanding exponential and logarithmic functions is key to grasping concepts like growth and decay models, which are vital in many scientific disciplines. You might find the resources at Khan Academy particularly helpful for additional explanations, practice problems, and interactive exercises on logarithms and related mathematical concepts. They offer a comprehensive approach that can solidify your understanding and help you tackle even more challenging problems.

For those looking for more advanced mathematical insights, resources like Brilliant.org offer interactive courses and problem-solving challenges that explore logarithms and other advanced mathematical topics in engaging ways. These platforms can provide a deeper, more intuitive understanding of how these mathematical tools are used to model and solve real-world problems.