Solving Simultaneous Equations: Logarithms And Exponentials

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Hey guys! Today, we're diving into the fascinating world of solving simultaneous equations, but with a twist! We're not just dealing with plain old algebra; we're tackling equations involving logarithms and exponentials. These types of problems might seem daunting at first, but don't worry, we'll break them down step by step. We'll explore different techniques and strategies to conquer these mathematical challenges. So, grab your pencils, and let's get started!

1. Solving Simultaneous Equations with Logarithms

Let's kick things off with our first set of equations. These involve the natural logarithm, denoted as "ln". Remember, the natural logarithm is simply the logarithm to the base e, where e is Euler's number (approximately 2.71828). Understanding the properties of logarithms is crucial for solving these equations. Key properties include:

  • ln(a) + ln(b) = ln(ab)
  • ln(a) - ln(b) = ln(a/b)
  • n ln(a) = ln(a^n)

Now, let's look at the equations themselves:

2 ln x + ln y = 1 + ln 5
ln 10x - ln y = 2 + ln 2

Step-by-Step Solution

Our goal is to find the values of x and y that satisfy both equations simultaneously. To do this, we'll use the properties of logarithms to simplify the equations and then employ techniques like substitution or elimination.

  1. Simplify using logarithm properties:

    • Equation 1: 2 ln x + ln y = ln(x^2) + ln(y) = ln(x^2y) = 1 + ln 5
    • Equation 2: ln 10x - ln y = ln(10x/y) = 2 + ln 2

  2. Remove the logarithms:

    To get rid of the logarithms, we'll exponentiate both sides of each equation using the base e. Remember that e^ln(a) = a.

    • From Equation 1: e^(ln(x^2y)) = e^(1 + ln 5) => x^2y = e * e^(ln 5) = 5e
    • From Equation 2: e^(ln(10x/y)) = e^(2 + ln 2) => 10x/y = e^2 * e^(ln 2) = 2e^2

  3. Solve the resulting equations:

    Now we have a system of two algebraic equations:

    • x^2y = 5e
    • 10x/y = 2e^2

    Let's solve for y in the second equation: y = 10x / (2e^2) = 5x/e^2

    Substitute this expression for y into the first equation:

    x^2 * (5x/e^2) = 5e => 5x^3/e^2 = 5e => x^3 = e^3 => x = e

  4. Find the value of y:

    Substitute x = e back into the equation for y:

    y = 5x/e^2 = 5e/e^2 = 5/e

  5. Final Solution:

    Therefore, the solution to the simultaneous equations is:

    • x = e
    • y = 5/e

Key Takeaways

  • The properties of logarithms are your best friends when solving these types of equations.
  • Exponentiating both sides is a powerful technique to eliminate logarithms.
  • Remember to check your solutions by substituting them back into the original equations.

2. Solving Simultaneous Equations with Exponentials

Now, let's move on to our second set of equations, which involve exponential functions. These functions have the variable in the exponent, and solving them often involves using logarithms (the inverse operation of exponentiation). Again, understanding the properties of exponents is essential here.

  • a^(m+n) = a^m * a^n
  • a^(m-n) = a^m / a^n
  • (a^m)^n = a^(mn)

Here are the equations we'll be tackling:

e^(3x + 4y) = 2e^2
e^(2x + y) = 8e^(x + 6y)

Step-by-Step Solution

Our strategy here will be to manipulate the equations, take logarithms of both sides, and then solve the resulting linear equations.

  1. Take the natural logarithm of both sides:

    This will help us bring the exponents down and create linear equations.

    • Equation 1: ln(e^(3x + 4y)) = ln(2e^2) => 3x + 4y = ln 2 + ln(e^2) = ln 2 + 2
    • Equation 2: ln(e^(2x + y)) = ln(8e^(x + 6y)) => 2x + y = ln 8 + ln(e^(x + 6y)) = ln 8 + x + 6y

  2. Simplify the equations:

    • Equation 1: 3x + 4y = ln 2 + 2
    • Equation 2: 2x + y = ln 8 + x + 6y => x - 5y = ln 8

    Note that ln 8 = ln(2^3) = 3 ln 2

    So, our simplified equations are:

    • 3x + 4y = ln 2 + 2
    • x - 5y = 3 ln 2

  3. Solve the linear equations:

    We can use either substitution or elimination. Let's use elimination. Multiply the second equation by -3:

    • 3x + 4y = ln 2 + 2
    • -3x + 15y = -9 ln 2

    Add the two equations:

    19y = -8 ln 2 + 2 => y = (2 - 8 ln 2) / 19

  4. Find the value of x:

    Substitute the value of y back into the equation x - 5y = 3 ln 2:

    x = 5y + 3 ln 2 = 5 * ((2 - 8 ln 2) / 19) + 3 ln 2

    x = (10 - 40 ln 2) / 19 + (57 ln 2) / 19 = (10 + 17 ln 2) / 19

  5. Final Solution:

    Therefore, the solution to the simultaneous equations is:

    • x = (10 + 17 ln 2) / 19
    • y = (2 - 8 ln 2) / 19

Key Takeaways

  • Taking logarithms of both sides is a standard technique for solving exponential equations.
  • Simplifying the resulting equations is crucial for making the problem manageable.
  • Don't be afraid of fractions or expressions involving logarithms in your final answer. Sometimes, that's just how it is!

3. General Strategies for Solving Simultaneous Equations

Whether you're dealing with logarithms, exponentials, or other types of functions, here are some general strategies that can help you solve simultaneous equations:

  • Simplify: Always start by simplifying the equations as much as possible. This might involve using algebraic identities, trigonometric identities, or properties of logarithms and exponents.
  • Isolate Variables: Try to isolate one variable in terms of the others. This will allow you to use substitution.
  • Substitution: If you can express one variable in terms of the others, substitute that expression into the other equations. This will reduce the number of variables and equations.
  • Elimination: Look for opportunities to eliminate variables by adding or subtracting multiples of the equations. This is particularly useful for linear equations.
  • Graphical Methods: For two-variable equations, you can sometimes graph the equations and find the points of intersection. This can be a helpful way to visualize the solutions.
  • Numerical Methods: If you can't find an exact solution, you can use numerical methods (like Newton's method) to approximate the solutions.
  • Check Your Solutions: Always check your solutions by substituting them back into the original equations. This will help you catch any errors.

Conclusion

Solving simultaneous equations involving logarithms and exponentials can be challenging, but with the right techniques and a good understanding of the properties of these functions, you can conquer these problems! Remember to practice regularly, and don't be afraid to ask for help when you get stuck. Keep up the great work, guys!