Math Problem Statement

solve for x 1.75e^(-x0.5) + 1.75e^(-x) + 1.75e^(-x1.5) + 101.75e^(-x*2)=86.3995

Solution

Let's solve the equation:

1.75e0.5x+1.75ex+1.75e1.5x+101.75e2x=86.39951.75e^{-0.5x} + 1.75e^{-x} + 1.75e^{-1.5x} + 101.75e^{-2x} = 86.3995

Step 1: Simplify the equation

First, let's factor out 1.751.75 from the first three terms:

1.75(e0.5x+ex+e1.5x)+101.75e2x=86.39951.75 \left( e^{-0.5x} + e^{-x} + e^{-1.5x} \right) + 101.75e^{-2x} = 86.3995

Next, divide the entire equation by 1.75 to further simplify:

e0.5x+ex+e1.5x+58.14e2x=49.37114e^{-0.5x} + e^{-x} + e^{-1.5x} + 58.14e^{-2x} = 49.37114

Step 2: Apply substitution

Let y=e0.5xy = e^{-0.5x}, so:

y2=ex,y3=e1.5x,y4=e2xy^2 = e^{-x}, \quad y^3 = e^{-1.5x}, \quad y^4 = e^{-2x}

Substitute into the equation:

y+y2+y3+58.14y4=49.37114y + y^2 + y^3 + 58.14y^4 = 49.37114

Step 3: Solve the polynomial equation

This is now a polynomial equation in yy:

58.14y4+y3+y2+y49.37114=058.14y^4 + y^3 + y^2 + y - 49.37114 = 0

To find the value of yy, this equation needs to be solved using numerical methods, such as Newton's method or by using a computational tool. Once yy is found, solve for xx using y=e0.5xy = e^{-0.5x}.

Let's solve this numerically to find the value of yy.The numerical solution for yy is approximately y0.9466y \approx 0.9466.

Step 4: Solve for xx

Recall that y=e0.5xy = e^{-0.5x}. So,

0.9466=e0.5x0.9466 = e^{-0.5x}

Take the natural logarithm on both sides:

ln(0.9466)=0.5x\ln(0.9466) = -0.5x

Solving for xx:

x=2ln(0.9466)x = -2 \ln(0.9466)

Now, let's compute the value of xx.The solution for xx is approximately x0.1098x \approx 0.1098.

Would you like any further details or have any questions? Here are some related problems you might find interesting:

  1. Solve a similar equation with different coefficients.
  2. Explore the solution method for a non-exponential polynomial equation.
  3. Analyze the behavior of the equation as one of the coefficients increases.
  4. Consider the case where the constant term on the right-hand side changes.
  5. Investigate the solutions for higher-order exponential equations.

Tip: When dealing with equations involving exponential terms, substituting variables can simplify the problem into a polynomial equation, making it easier to solve.

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Math Problem Analysis

Mathematical Concepts

Exponential Equations
Polynomial Substitution
Numerical Methods

Formulas

y = e^{-0.5x}
58.14y^4 + y^3 + y^2 + y - 49.37114 = 0
x = -2 * ln(0.9466)

Theorems

Logarithmic Identity
Numerical Solution of Polynomials

Suitable Grade Level

College Level (Advanced Algebra/Pre-Calculus)