Solving 9^x - 3^(x+1) + 8/9 = 0 And Probability Of Two Equal Digits In Three-Digit Numbers

Jun 27, 2025 by ADMIN 91 views

Solve the equation 9^x - 3^(x+1) + 8/9 = 0 in R. Determine the probability that a randomly selected three-digit natural number has exactly two digits equal.

Solving the Equation 9^x - 3^(x+1) + 8/9 = 0 and Probability of Two Equal Digits in Three-Digit Numbers

Introduction

This article delves into two distinct mathematical problems. First, we will solve the exponential equation 9^x - 3^(x+1) + 8/9 = 0 in the set of real numbers. This involves algebraic manipulation and the application of exponential properties to find the solution set. Second, we will tackle a probability problem, determining the probability that a randomly chosen three-digit natural number has exactly two digits that are equal. This requires combinatorial reasoning and a clear understanding of probability principles. These types of problems are fundamental in mathematics and often appear in various competitive exams and real-world applications. Understanding these concepts is crucial for anyone looking to enhance their mathematical skills and problem-solving abilities.

Solving the Exponential Equation 9^x - 3^(x+1) + 8/9 = 0

Transforming the Equation

To solve the exponential equation 9^x - 3^(x+1) + 8/9 = 0, we first need to transform it into a more manageable form. Notice that 9^x can be expressed as (3²⁾x, which is equal to (3^x)2. Also, 3^(x+1) can be rewritten as 3 * 3^x. Substituting these expressions into the original equation, we get: (3^x)2 - 3 * 3^x + 8/9 = 0. This transformation sets the stage for a substitution that will simplify the equation into a quadratic form. Recognizing the relationship between the exponential terms allows us to apply standard algebraic techniques to find the solution.

Substitution

Now, let's make a substitution to simplify the equation further. Let y = 3^x. Substituting y into the transformed equation, we obtain a quadratic equation in terms of y: y^2 - 3y + 8/9 = 0. This quadratic equation is much easier to solve than the original exponential equation. By using a simple substitution, we have converted the problem into a familiar algebraic form. This step is crucial in solving many complex equations, where recognizing patterns and making appropriate substitutions can significantly simplify the problem.

Solving the Quadratic Equation

To solve the quadratic equation y^2 - 3y + 8/9 = 0, we can use the quadratic formula or try to factor the equation. Let's use the quadratic formula, which states that for an equation of the form ay^2 + by + c = 0, the solutions for y are given by: y = [-b ± √(b^2 - 4ac)] / (2a). In our case, a = 1, b = -3, and c = 8/9. Plugging these values into the formula, we get: y = [3 ± √((-3)^2 - 4 * 1 * (8/9))] / (2 * 1). Simplifying the expression under the square root, we have: y = [3 ± √(9 - 32/9)] / 2. Further simplification gives: y = [3 ± √(49/9)] / 2. Thus, y = [3 ± (7/3)] / 2. This gives us two possible values for y.

Finding the Values of y

We have two possible values for y: y1 = (3 + 7/3) / 2 and y2 = (3 - 7/3) / 2. Let's calculate these values. For y1: y1 = (3 + 7/3) / 2 = (16/3) / 2 = 8/3. For y2: y2 = (3 - 7/3) / 2 = (2/3) / 2 = 1/3. So, the two values of y are 8/3 and 1/3. These values are the solutions to the quadratic equation we obtained after the substitution. Now, we need to substitute back to find the values of x.

Substituting Back to Find x

Now that we have the values of y, we need to substitute back to find the corresponding values of x. Recall that we made the substitution y = 3^x. So, we have two equations to solve: 3^x = 8/3 and 3^x = 1/3. For the first equation, 3^x = 8/3, we can take the logarithm base 3 of both sides: x = log₃(8/3). For the second equation, 3^x = 1/3, we can rewrite 1/3 as 3^(-1), so we have: 3^x = 3^(-1). Thus, x = -1. These are the solutions for x that satisfy the original exponential equation.

Final Solutions for x

We have found two solutions for x: x = log₃(8/3) and x = -1. These are the real solutions to the original equation 9^x - 3^(x+1) + 8/9 = 0. The solution set can be written as {-1, log₃(8/3)}. To verify these solutions, we can substitute them back into the original equation and check if they satisfy it. The process of solving this exponential equation demonstrates the importance of algebraic manipulation, substitution, and the use of logarithms to find the solutions.

Determining the Probability of Two Equal Digits in Three-Digit Numbers

Understanding the Problem

Now, let's shift our focus to the probability problem. We want to find the probability that a randomly chosen three-digit natural number has exactly two digits that are equal. This problem involves understanding the structure of three-digit numbers and applying principles of combinatorics to count the favorable and total outcomes. It's essential to consider the constraints on the digits (e.g., the first digit cannot be zero) and systematically count the possibilities.

Counting Total Three-Digit Numbers

To calculate the probability, we first need to determine the total number of three-digit natural numbers. The smallest three-digit number is 100, and the largest is 999. Therefore, the total number of three-digit numbers is 999 - 100 + 1 = 900. This will be the denominator in our probability calculation. Understanding the range of numbers we are considering is the first step in solving any probability problem.

Counting Favorable Outcomes

Now, we need to count the number of three-digit numbers with exactly two digits that are equal. This is a bit more complex and requires careful consideration of different cases. Let's break it down into cases based on the positions of the equal digits. We have three positions for the digits: hundreds, tens, and units. We need to count the numbers where exactly two of these positions have the same digit.

Case 1: Hundreds and Tens Digits are Equal

In the first case, let's consider the situation where the hundreds and tens digits are equal. The hundreds digit can be any digit from 1 to 9 (it cannot be 0). So, there are 9 choices for the hundreds digit. Since the tens digit must be the same as the hundreds digit, there is only 1 choice for the tens digit. The units digit must be different from the hundreds digit, so there are 9 choices for the units digit (0 to 9, excluding the digit used for hundreds and tens). Therefore, the number of three-digit numbers in this case is 9 * 1 * 9 = 81.

Case 2: Hundreds and Units Digits are Equal

In the second case, the hundreds and units digits are equal. Again, the hundreds digit can be any digit from 1 to 9, so there are 9 choices. Since the units digit must be the same, there is only 1 choice for the units digit. The tens digit must be different from the hundreds digit, so there are 9 choices for the tens digit (0 to 9, excluding the digit used for hundreds and units). Therefore, the number of three-digit numbers in this case is also 9 * 9 * 1 = 81.

Case 3: Tens and Units Digits are Equal

Finally, in the third case, the tens and units digits are equal. The hundreds digit can be any digit from 1 to 9, so there are 9 choices. The tens and units digits must be the same, and they must be different from the hundreds digit. Since the tens digit can be 0, there are 9 choices for the tens digit (0 to 9, excluding the digit used for hundreds). Therefore, the number of three-digit numbers in this case is 9 * 9 * 1 = 81.

Total Favorable Outcomes

Now, we sum the favorable outcomes from all three cases. The total number of three-digit numbers with exactly two digits equal is 81 + 81 + 81 = 243. This is the number of outcomes that satisfy our condition.

Calculating the Probability

Finally, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes. The probability P is given by: P = (Number of favorable outcomes) / (Total number of outcomes) = 243 / 900. We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 27. So, P = (243 / 27) / (900 / 27) = 9 / (100/3) = 27/100. Thus, the probability that a randomly chosen three-digit natural number has exactly two digits that are equal is 27/100 or 0.27.

Conclusion

In this article, we have tackled two different mathematical problems. First, we successfully solved the exponential equation 9^x - 3^(x+1) + 8/9 = 0, finding the solutions x = -1 and x = log₃(8/3). This required transforming the equation, making a substitution, solving a quadratic equation, and substituting back to find the values of x. Second, we determined the probability that a randomly chosen three-digit natural number has exactly two digits that are equal, which is 27/100. This involved counting the total number of three-digit numbers and the number of such numbers with exactly two equal digits, then calculating the probability. Both problems highlight the importance of understanding mathematical concepts and applying problem-solving techniques to find solutions. These skills are valuable in various fields and can enhance one's mathematical aptitude.