Evaluate Log₇98: A Step-by-Step Guide With Log Properties

Hey guys! Let's break down this logarithm problem step by step. We're given that ${\log _7 2 \approx 0.356}$, and our mission is to figure out ${\log _7 98}$. To nail this, we'll be using some cool properties of logarithms, especially the product property. So, let’s dive in and make this log problem a piece of cake!

Understanding the Problem

So, the big question here is: How can we evaluate ${\log _7 98}$ when we know that ${\log _7 2 \approx 0.356}$? This is where the magic of logarithmic properties comes into play. Logarithms can seem intimidating, but they're really just a different way of thinking about exponents. Remember that ${\log _b a = c}$ is the same as saying ${b^c = a}$. Our goal is to rewrite ${\log _7 98}$ in a way that involves ${\log _7 2}$, which we already have a value for. To do this, we’ll need to factor 98 and use the properties of logarithms to break it down into simpler terms. Factoring 98 will help us express it as a product of numbers, one of which will hopefully be 2, given our known logarithm. This is where the product property becomes super handy.

Why are we doing this? Well, logarithms have these nifty rules that let us simplify complex expressions. One of these is the product property, which we’ll get into shortly. By rewriting 98 as a product, we can split the logarithm into a sum of logarithms, making it easier to solve. Think of it like breaking a big problem into smaller, more manageable chunks. This is a common strategy in math, and it’s exactly what we’re going to do here. Now, let's explore how we can use this product property to rewrite the expression and get closer to our solution. We need to really understand how 98 can be factored in a way that uses the number 2, as we already know the ${\log _7}$ of 2. So, stick with me, and let’s make some logarithmic magic happen!

Using the Product Property

The key to cracking this problem is the product property of logarithms. This property is super useful and states that ${\log _b (mn) = \log _b m + \log _b n}$. In plain English, the logarithm of a product is equal to the sum of the logarithms of the individual factors. This is exactly what we need to break down ${\log _7 98}$.

So, how do we apply this to our problem? First, we need to express 98 as a product of its factors. We know that 98 can be written as ${2 \times 49}$. Why is this helpful? Because we know ${\log _7 2}$, and 49 is a power of 7! Specifically, ${49 = 7^2}$. This is great because we can easily find ${\log _7 49}$. Now we can rewrite ${\log _7 98}$ using our factors:

${\log _7 98 = \log _7 (2 \times 49)}$

Now, using the product property, we can split this into two separate logarithms:

${\log _7 (2 \times 49) = \log _7 2 + \log _7 49}$

This is a major step forward! We've successfully rewritten ${\log _7 98}$ as a sum of two logarithms. We know the value of ${\log _7 2}$, and ${\log _7 49}$ is something we can easily calculate because 49 is a power of 7. Let's look at the options given in the question and see which one matches our rewritten expression. Option A, ${7 \log 2 + 7 \log 49}$, is incorrect because it multiplies each logarithm by 7, which isn’t what the product property tells us to do. Option B, ${\log 7 + \log 2 + \log 49}$, is also incorrect because it includes ${\log 7}$, which isn’t part of our rewritten expression. Option C, ${\log _7 2 + \log _7 49}$, matches perfectly! We’ve correctly used the product property to rewrite the original logarithm. Next, we'll figure out the value of ${\log _7 49}$ to get closer to our final answer.

Calculating log₇49

Alright, let's tackle the next part: calculating ${\log _7 49}$. This might seem tricky at first, but it’s actually quite straightforward once you remember what a logarithm represents. Recall that ${\log _b a = c}$ is just another way of saying ${b^c = a}$. So, in our case, we're asking ourselves, “To what power must we raise 7 to get 49?”

Think about it: 49 is a perfect square, and it’s specifically the square of 7. That is, ${49 = 7^2}$. This makes our job super easy! If ${7^2 = 49}$, then, by definition, ${\log _7 49 = 2}$. See? No complex calculations needed; it’s all about recognizing the relationship between the base (7) and the number (49).

So, we’ve determined that ${\log _7 49 = 2}$. This is a crucial piece of the puzzle. We now know both ${\log _7 2}$ (which was given as approximately 0.356) and ${\log _7 49}$. We're just one step away from finding ${\log _7 98}$. Remember, we used the product property to rewrite ${\log _7 98}$ as ${\log _7 2 + \log _7 49}$. Now that we have the values for both of these logarithms, we can simply add them together. Let’s do that in the next section and get our final answer!

Final Evaluation

Okay, we're in the home stretch now! We've done the hard work of rewriting the original logarithm using the product property and figuring out the value of ${\log _7 49}$. To recap, we know:

• ${\log _7 2 \approx 0.356}$
• ${\log _7 49 = 2}$

And we rewrote ${\log _7 98}$ as:

${\log _7 98 = \log _7 2 + \log _7 49}$

Now it’s just a matter of plugging in the values we know:

${\log _7 98 \approx 0.356 + 2}$

Adding these together is simple:

${\log _7 98 \approx 2.356}$

So, there you have it! We've successfully evaluated ${\log _7 98}$. By using the product property of logarithms and our knowledge of powers, we were able to break down a seemingly complex problem into manageable parts. This approach highlights the power of understanding logarithmic properties – they can turn daunting problems into simple arithmetic. Remember, the key was to rewrite 98 as a product of 2 and 49, then apply the product property to split the logarithm, and finally, use the definition of logarithms to evaluate ${\log _7 49}$. This step-by-step method is a great way to tackle similar problems in the future. Next time you see a logarithm problem, think about how you can break it down using these properties – you might be surprised at how easy it becomes!

Conclusion

In conclusion, guys, we've successfully navigated the world of logarithms and evaluated ${\log _7 98}$ using the product property. The key takeaways here are:

1. Product Property: Remember that ${\log _b (mn) = \log _b m + \log _b n}$. This property is your best friend when dealing with logarithms of products.
2. Factoring: Breaking down the number inside the logarithm into its factors can help you simplify the expression.
3. Definition of Logarithms: Don't forget what a logarithm actually means. ${\log _b a = c}$ is just another way of saying ${b^c = a}$. This is crucial for evaluating logarithms like ${\log _7 49}$.

By combining these concepts, we turned a potentially tricky problem into a straightforward calculation. So, keep practicing with these logarithmic properties, and you'll become a pro in no time! Now you can confidently say you know how to evaluate logarithms using their properties. Keep up the great work, and remember, math can be fun when you break it down step by step!