Calculate Log Base 4 Of 7 Using Log Properties

Hey guys! Let's dive into a cool math problem involving logarithms. We're given that $\log _4 3 \approx 0.792$ and $\log _4 21 \approx 2.196$, and our mission is to find the value of $\log _4 7$. Sounds like a fun puzzle, right? Let's break it down step by step.

Understanding Logarithms

Before we jump into solving the problem, let's quickly refresh what logarithms are all about. A logarithm is essentially the inverse operation to exponentiation. When we write $\log_b a = c$, it means that $b^c = a$. In simpler terms, the logarithm tells you what power you need to raise the base ($b$) to in order to get a certain number ($a$).

Logarithms have some really neat properties that make them super useful in simplifying complex calculations. One of the most important properties is the logarithm of a product, which states that the logarithm of the product of two numbers is equal to the sum of their logarithms. Mathematically, this is expressed as:

$\log_b (xy) = \log_b x + \log_b y$

Another useful property is the logarithm of a quotient, which states that the logarithm of the quotient of two numbers is equal to the difference of their logarithms. Mathematically, this is expressed as:

$\log_b (x/y) = \log_b x - \log_b y$

These properties will be our best friends as we tackle the problem at hand. We will leverage these rules to manipulate the given logarithmic expressions and isolate the value we're trying to find.

Solving for $\log _4 7$

Okay, let's get back to our problem. We know that $\log _4 3 \approx 0.792$ and $\log _4 21 \approx 2.196$. Our goal is to find $\log _4 7$. The key here is to recognize the relationship between 21, 3, and 7. Notice that 21 can be expressed as the product of 3 and 7, i.e., $21 = 3 \times 7$. This is where the logarithm properties come into play!

Using the logarithm of a product property, we can write:

$\log _4 21 = \log _4 (3 \times 7)$

And then, we can expand this using the property:

$\log _4 21 = \log _4 3 + \log _4 7$

Now, we can substitute the given values:

$2.196 = 0.792 + \log _4 7$

To find $\log _4 7$, we simply subtract 0.792 from both sides of the equation:

$\log _4 7 = 2.196 - 0.792$

$\log _4 7 = 1.404$

So, there you have it! We found that $\log _4 7 \approx 1.404$. This matches option A. Isn't it satisfying when everything clicks into place?

Why This Works: A Deeper Dive

You might be wondering, why does this logarithm property work? Let's think about it in terms of exponents. Suppose we have:

$\log_b x = m$ which means $b^m = x$ $\log_b y = n$ which means $b^n = y$

Then, when we multiply $x$ and $y$, we get:

$x \times y = b^m \times b^n = b^{m+n}$

Now, if we take the logarithm base $b$ of both sides:

$\log_b (xy) = \log_b (b^{m+n})$

Using the property that $\log_b (b^k) = k$, we get:

$\log_b (xy) = m + n$

And since $m = \log_b x$ and $n = \log_b y$, we have:

$\log_b (xy) = \log_b x + \log_b y$

This shows why the logarithm of a product is the sum of the logarithms. It's all rooted in the fundamental relationship between logarithms and exponents.

Applying This Knowledge

Understanding these logarithm properties isn't just about solving textbook problems. They pop up in various real-world applications. For instance, in computer science, logarithms are used in analyzing the complexity of algorithms. In physics, they appear in formulas related to entropy and information theory. Even in finance, logarithmic scales are sometimes used to represent stock prices or economic growth.

So, the next time you encounter a situation involving exponential growth or decay, remember the power of logarithms. They can often simplify complex relationships and make calculations much more manageable. Keep practicing, and you'll become a logarithm pro in no time!

Practice Problems

Want to test your understanding? Here are a couple of practice problems you can try:

1. Given $\log _2 5 \approx 2.322$ and $\log _2 3 \approx 1.585$, find $\log _2 15$.
2. Given $\log _5 10 \approx 1.431$ and $\log _5 2 \approx 0.431$, find $\log _5 5$.

Try solving these using the logarithm properties we discussed. The answers are at the end of this article, but try to work them out on your own first!

Common Mistakes to Avoid

When working with logarithms, it's easy to make a few common mistakes. Here are some things to watch out for:

• Forgetting the Base: Always remember to write the base of the logarithm. $\log x$ without a base is ambiguous and can lead to errors.
• Incorrectly Applying Properties: Make sure you understand the logarithm properties correctly before applying them. For example, $\log (x + y)$ is not equal to $\log x + \log y$.
• Dividing Inside Logarithms: Avoid dividing inside logarithms unless you are using the quotient rule properly. $\frac{\log x}{\log y}$ is not the same as $\log \frac{x}{y}$.
• Confusing Logarithms with Exponents: Remember that logarithms are the inverse of exponents. Keep the definitions clear in your mind.

By being mindful of these potential pitfalls, you can avoid making errors and solve logarithm problems with confidence.

Conclusion

Logarithms might seem a bit abstract at first, but with practice and a solid understanding of their properties, they become a powerful tool in your mathematical arsenal. In this article, we tackled the problem of finding $\log _4 7$ given $\log _4 3$ and $\log _4 21$. We used the logarithm of a product property to break down the problem and find the solution. Keep exploring, keep practicing, and you'll become a logarithm master in no time!

Answers to Practice Problems:

1. $\log _2 15 = \log _2 (5 \times 3) = \log _2 5 + \log _2 3 \approx 2.322 + 1.585 = 3.907$
2. $\log _5 5 = \log _5 (10/2) = \log _5 10 - \log _5 2 \approx 1.431 - 0.431 = 1$ (Alternatively, $\log_5 5 = 1$ directly, since $5^1 = 5$)

Hope you guys found this helpful and happy calculating!