Solving Logarithmic And Calendar Math Problems

Hey guys! Let's dive into the fascinating world of logarithms, and specifically, let's tackle a problem where we need to find the value of log₁₀ 4 + log₁₀ 16, given that log₁₀ 8 = 0.9031. This problem is a classic example of how we can manipulate logarithmic expressions using their fundamental properties. So, buckle up, and let's get started!

Understanding the Fundamentals of Logarithms

Before we jump into solving the problem, let's quickly brush up on the basics of logarithms. A logarithm is essentially the inverse operation of exponentiation. In simpler terms, if we have an equation like bˣ = y, then we can express this in logarithmic form as log_b y = x. Here, 'b' is the base of the logarithm, 'y' is the argument, and 'x' is the exponent to which we need to raise the base 'b' to get 'y'.

In our problem, we're dealing with base-10 logarithms, which are also known as common logarithms. This means that the base 'b' is 10. So, log₁₀ y = x implies that 10ˣ = y. The beauty of logarithms lies in their ability to simplify complex calculations involving multiplication, division, and exponentiation. They allow us to transform these operations into simpler addition and subtraction, making them incredibly useful in various fields like science, engineering, and finance.

Leveraging Logarithmic Properties to Solve the Problem

Now, let's get back to the problem at hand: finding the value of log₁₀ 4 + log₁₀ 16, given that log₁₀ 8 = 0.9031. To solve this, we'll need to utilize some key properties of logarithms. The most important property for this problem is the product rule, which states that the logarithm of a product is equal to the sum of the logarithms of the individual factors. Mathematically, this can be expressed as:

log_b (mn) = log_b m + log_b n

Another crucial property we'll use is the power rule, which states that the logarithm of a number raised to a power is equal to the power multiplied by the logarithm of the number. In mathematical notation:

log_b (mⁿ) = n log_b m

With these properties in our arsenal, we can now break down the problem into smaller, manageable steps. First, let's express 4 and 16 as powers of 2, since we know the value of log₁₀ 8, and 8 is also a power of 2. We can write 4 as 2² and 16 as 2⁴. Substituting these into our expression, we get:

log₁₀ 4 + log₁₀ 16 = log₁₀ (2²) + log₁₀ (2⁴)

Next, we apply the power rule to both terms:

log₁₀ (2²) + log₁₀ (2⁴) = 2 log₁₀ 2 + 4 log₁₀ 2

Now, we can combine the terms since they both have log₁₀ 2:

2 log₁₀ 2 + 4 log₁₀ 2 = 6 log₁₀ 2

We're almost there! We now have the expression 6 log₁₀ 2, but we know the value of log₁₀ 8, not log₁₀ 2. So, we need to find a way to relate these two. We can express 8 as 2³, and using the power rule again, we have:

log₁₀ 8 = log₁₀ (2³) = 3 log₁₀ 2

We are given that log₁₀ 8 = 0.9031, so:

3 log₁₀ 2 = 0.9031

Dividing both sides by 3, we get:

log₁₀ 2 = 0.9031 / 3 = 0.3010333...

Now we can substitute this value back into our expression 6 log₁₀ 2:

6 log₁₀ 2 = 6 * 0.3010333... = 1.8062

Therefore, the value of log₁₀ 4 + log₁₀ 16 is approximately 1.8062.

A Quick Recap

Let's quickly recap the steps we took to solve this problem:

1. We expressed 4 and 16 as powers of 2.
2. We applied the power rule of logarithms to simplify the expression.
3. We used the given value of log₁₀ 8 to find the value of log₁₀ 2.
4. We substituted the value of log₁₀ 2 back into our expression to get the final answer.

By understanding and applying the fundamental properties of logarithms, we were able to solve this problem efficiently and accurately. Logarithms might seem intimidating at first, but with practice and a solid grasp of their properties, you'll be able to conquer any logarithmic challenge!

Hey everyone! Let's tackle two interesting problems today. The first one is a fun calendar puzzle, and the second one involves solving a logarithmic equation. Both problems require a bit of logical thinking and a good understanding of mathematical principles. So, let's put on our thinking caps and get started!

i) Decoding the Days: What Day Will It Be After 200 Days?

This first part is a classic problem that combines calendar knowledge with basic arithmetic. We're given that today is Tuesday, and we need to figure out what day it will be after 200 days. The key to solving this lies in understanding the cyclical nature of days in a week. There are 7 days in a week, and after every 7 days, the cycle repeats itself. This means that we can use the concept of remainders to determine the day of the week after a certain number of days.

The Cyclical Nature of Days and Remainders

Think of it like this: if today is Tuesday, then 7 days from now it will also be Tuesday. Similarly, 14 days from now, it will be Tuesday again, and so on. This pattern continues because the days of the week form a cycle. To find out what day it will be after 200 days, we need to figure out how many full weeks are in 200 days and what the remainder is. This remainder will tell us how many days to count forward from Tuesday.

To find the number of full weeks, we simply divide 200 by 7:

200 ÷ 7 = 28 with a remainder of 4

This tells us that there are 28 full weeks in 200 days, and there are 4 days left over. The 28 full weeks don't change the day of the week, as we end up back on Tuesday after each full week. However, the remainder of 4 is crucial. It means that we need to count 4 days forward from Tuesday to find the day of the week after 200 days.

Counting Forward from Tuesday

Let's count those 4 days: Wednesday, Thursday, Friday, Saturday. So, 200 days from Tuesday, it will be Saturday. This problem highlights the importance of recognizing patterns and using remainders to solve cyclical problems. It's a simple yet elegant way to navigate the calendar and predict days in the future.

ii) Unraveling Logarithmic Equations: Finding the Value of x

Now, let's shift gears and dive into the world of logarithmic equations. We're given the equation log(5x - 4) = log(x + 1) + log 4, and our mission is to find the value of x that satisfies this equation. To do this, we'll need to leverage the properties of logarithms, just like we did in the previous problem. The key here is to simplify the equation using logarithmic rules and then solve for x using algebraic techniques.

The Power of Logarithmic Properties

Remember those logarithmic properties we discussed earlier? They're going to be our best friends in solving this equation. In particular, we'll use the product rule, which states that the logarithm of a product is equal to the sum of the logarithms of the individual factors:

log_b (mn) = log_b m + log_b n

We'll also use the fact that if log_b m = log_b n, then m = n. This is a fundamental property that allows us to eliminate logarithms from an equation and solve for the unknown variable.

Simplifying the Equation

Let's apply the product rule to the right side of our equation:

log(5x - 4) = log(x + 1) + log 4

log(5x - 4) = log[4(x + 1)]

Now we have logarithms on both sides of the equation. Since the logarithms have the same base (base 10, since it's a common logarithm), we can equate the arguments:

5x - 4 = 4(x + 1)

Solving for x Algebraically

We've successfully transformed our logarithmic equation into a simple algebraic equation. Now, it's just a matter of solving for x. Let's distribute the 4 on the right side:

5x - 4 = 4x + 4

Subtract 4x from both sides:

x - 4 = 4

Add 4 to both sides:

x = 8

Therefore, the value of x that satisfies the equation log(5x - 4) = log(x + 1) + log 4 is 8.

Verifying the Solution

It's always a good idea to verify our solution by plugging it back into the original equation. Let's substitute x = 8 into the equation:

log(5(8) - 4) = log(8 + 1) + log 4

log(40 - 4) = log 9 + log 4

log 36 = log (9 * 4)

log 36 = log 36

Our solution checks out! This confirms that x = 8 is indeed the correct answer.

A Quick Review

Let's recap the steps we took to solve the logarithmic equation:

1. We applied the product rule of logarithms to simplify the equation.
2. We equated the arguments of the logarithms since they had the same base.
3. We solved the resulting algebraic equation for x.
4. We verified our solution by plugging it back into the original equation.

Logarithmic equations might seem tricky at first, but by mastering the properties of logarithms and practicing algebraic manipulation, you can confidently solve these types of problems.

This article explores mathematical problems, providing detailed solutions and explanations. From logarithmic calculations to calendar-based questions, we aim to break down complex concepts into digestible steps. Whether you're a student looking to improve your math skills or simply someone who enjoys problem-solving, this article offers valuable insights and techniques. Join us as we unravel these challenges and enhance your understanding of mathematics.