Math Problem Statement

  1. Find tanhx ,if 6sinhx +2 coshx + 7 =0

Solution

To find tanh(x)\tanh(x), given the equation:

6sinh(x)+2cosh(x)+7=06 \sinh(x) + 2 \cosh(x) + 7 = 0

Step 1: Express the hyperbolic functions in terms of exponentials.

We know that:

  • sinh(x)=exex2\sinh(x) = \frac{e^x - e^{-x}}{2}
  • cosh(x)=ex+ex2\cosh(x) = \frac{e^x + e^{-x}}{2}

Substituting these into the equation:

6(exex2)+2(ex+ex2)+7=06 \left(\frac{e^x - e^{-x}}{2}\right) + 2 \left(\frac{e^x + e^{-x}}{2}\right) + 7 = 0

Step 2: Simplify the equation.

Simplifying the terms:

3(exex)+(ex+ex)+7=03(e^x - e^{-x}) + (e^x + e^{-x}) + 7 = 0

Combine like terms:

(3ex+ex)(3exex)+7=0(3e^x + e^x) - (3e^{-x} - e^{-x}) + 7 = 0

This simplifies to:

4ex2ex+7=04e^x - 2e^{-x} + 7 = 0

Step 3: Solve for exe^x or exe^{-x}.

Let's solve this equation for exe^x.

  1. Move the constant to the other side:

4ex2ex=74e^x - 2e^{-x} = -7

This is a transcendental equation and typically requires solving through a substitution or numerical method. We can proceed by assuming a substitution or numerical approach to find a precise solution. Let's solve it numerically to determine the value of xx and then find tanh(x)\tanh(x).

Let me calculate this for you.The solutions for xx are complex expressions involving logarithms. However, the values of tanh(x)\tanh(x) for the solutions are:

  1. tanh(log(72372))\tanh\left(\log\left(\frac{7}{2} - \frac{\sqrt{37}}{2}\right)\right)
  2. tanh(log(372+72))\tanh\left(\log\left(\frac{\sqrt{37}}{2} + \frac{7}{2}\right)\right)

These expressions represent the hyperbolic tangent of xx for the two solutions. They can be further evaluated numerically if needed.

Would you like the numerical values of tanh(x)\tanh(x), or do you need further clarification?

Here are five related questions for deeper understanding:

  1. How are hyperbolic functions related to exponential functions?
  2. What is the geometric interpretation of tanh(x)\tanh(x)?
  3. How can transcendental equations be solved numerically?
  4. What are the differences between trigonometric and hyperbolic functions?
  5. Can we use a graphical method to estimate solutions to hyperbolic equations?

Tip: Hyperbolic functions are analogs of trigonometric functions, but they are based on exponential functions rather than circular functions.

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Math Problem Analysis

Mathematical Concepts

Hyperbolic functions
Exponential functions
Transcendental equations

Formulas

\(\sinh(x) = \frac{e^x - e^{-x}}{2}\)
\(\cosh(x) = \frac{e^x + e^{-x}}{2}\)
Transcendental equation solving methods

Theorems

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Suitable Grade Level

Advanced Mathematics