Gauss Elimination And Logarithm Problems Solutions

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Okay guys, let's dive into solving a system of linear equations using the Gauss Elimination Method. This is a super useful technique in linear algebra, and once you get the hang of it, you'll be solving systems like a pro!

Understanding the Gauss Elimination Method

The Gauss Elimination method, at its heart, is a systematic approach to transform a system of linear equations into an equivalent, simpler system that can be easily solved. This "simpler system" is usually in what we call row-echelon form or reduced row-echelon form. Think of it like this: we're organizing the equations in a way that makes the solution pop out. The method involves performing elementary row operations on the augmented matrix of the system. These operations are crucial and include:

  1. Swapping two rows: You can interchange any two equations in the system without changing the solution. It's like reordering your shopping list – you still have the same items.
  2. Multiplying a row by a non-zero constant: If you multiply an entire equation by a number (except zero), the solutions remain the same. Imagine scaling a recipe up or down; the proportions are the same.
  3. Adding a multiple of one row to another: This is where the magic happens. By adding a multiple of one equation to another, we can eliminate variables strategically.

The main goal is to get the matrix into an upper triangular form (row-echelon form), where all entries below the main diagonal are zeros. Once we have this form, we can easily use back-substitution to find the values of the variables.

Applying Gauss Elimination to the System

Let's apply this method to the given system of equations:

3x + 2y - 5z = -8
2x + 3y + 6z = 36
x - y + 6z = 7

First, we represent this system as an augmented matrix:

[ 3  2 -5 | -8 ]
[ 2  3  6 | 36 ]
[ 1 -1  6 |  7 ]

Our goal is to transform this matrix into row-echelon form. We'll do this step-by-step.

Step 1: Get a '1' in the top-left corner

It's often easiest to start by getting a '1' in the first row, first column position. We can achieve this by swapping Row 1 and Row 3:

[ 1 -1  6 |  7 ]
[ 2  3  6 | 36 ]
[ 3  2 -5 | -8 ]

Step 2: Eliminate 'x' from the second and third equations

Now, we want to eliminate the 'x' term from the second and third rows. To do this:

  • Subtract 2 times Row 1 from Row 2 (R2 = R2 - 2R1)
  • Subtract 3 times Row 1 from Row 3 (R3 = R3 - 3R1)

This gives us:

[ 1 -1  6 |  7 ]
[ 0  5 -6 | 22 ]
[ 0  5 -23 | -29 ]

Step 3: Eliminate 'y' from the third equation

Next, we eliminate the 'y' term from the third row. Subtract Row 2 from Row 3 (R3 = R3 - R2):

[ 1 -1  6 |  7 ]
[ 0  5 -6 | 22 ]
[ 0  0 -17 | -51 ]

Step 4: Convert to equations and solve

Now we have the matrix in row-echelon form. Let's convert it back into equations:

x - y + 6z = 7
5y - 6z = 22
-17z = -51

From the third equation, we can directly find 'z':

z = -51 / -17 = 3

Now we use back-substitution. Plug 'z = 3' into the second equation:

5y - 6(3) = 22
5y - 18 = 22
5y = 40
y = 8

Finally, plug 'y = 8' and 'z = 3' into the first equation:

x - 8 + 6(3) = 7
x - 8 + 18 = 7
x + 10 = 7
x = -3

So, the solution is x = -3, y = 8, and z = 3. We've successfully used Gauss Elimination to solve the system!

Now, let's tackle a logarithm problem. We're given that log₁₀2 = 0.3010, and we need to find the value of log₈25. This is where understanding the properties of logarithms becomes super handy.

Understanding Logarithm Properties

The key to solving this problem lies in using the change of base formula and understanding basic logarithm properties.

  • Change of Base Formula: This formula allows us to convert a logarithm from one base to another. It states:

    logₐb = logₓb / logₓa

    where 'a', 'b', and 'x' are positive numbers and 'a' and 'x' are not equal to 1. This formula is gold because it lets us switch to a base we know something about (like base 10 in this case).

  • Power Rule: This rule says that the logarithm of a number raised to a power is equal to the power times the logarithm of the number:

    logₐ(b^c) = c * logₐb

    This is super helpful for simplifying expressions.

Solving the Logarithm Problem

Let's use these properties to find log₈25. First, we'll use the change of base formula to convert log₈25 to base 10:

log₈25 = log₁₀25 / log₁₀8

Now, we can rewrite 25 as 5² and 8 as 2³:

log₈25 = log₁₀(5²) / log₁₀(2³)

Using the power rule, we can bring the exponents down:

log₈25 = (2 * log₁₀5) / (3 * log₁₀2)

We know log₁₀2 = 0.3010, but we need to find log₁₀5. Remember that 5 = 10 / 2. So, we can use the logarithm property log(a/b) = log(a) - log(b):

log₁₀5 = log₁₀(10/2) = log₁₀10 - log₁₀2

Since log₁₀10 = 1:

log₁₀5 = 1 - log₁₀2 = 1 - 0.3010 = 0.6990

Now we have all the pieces! Plug the values back into our equation:

log₈25 = (2 * 0.6990) / (3 * 0.3010)
log₈25 = 1.398 / 0.903
log₈25 ≈ 1.548

So, the value of log₈25 is approximately 1.548. We cracked it!

Conclusion

We've tackled two important mathematical problems: solving a system of linear equations using Gauss Elimination and evaluating a logarithm using the change of base formula and logarithm properties. Both of these techniques are fundamental in mathematics, and mastering them will definitely boost your problem-solving skills. Keep practicing, guys, and you'll become math whizzes in no time!