Solve by using English language

Recall that $\cosh x = \frac{e^x + e^{-x}}{2}$ and $\sinh x = \frac{e^x - e^{-x}}{2}$.

$$2 \left(\frac{e^x + e^{-x}}{2}\right) + 3 \left(\frac{e^x - e^{-x}}{2}\right) = 3$$

$$(e^x + e^{-x}) + \frac{3}{2}(e^x - e^{-x}) = 3$$

Multiply the entire equation by 2 to eliminate the fraction:

$$2(e^x + e^{-x}) + 3(e^x - e^{-x}) = 6$$

$$2e^x + 2e^{-x} + 3e^x - 3e^{-x} = 6$$ $$(2e^x + 3e^x) + (2e^{-x} - 3e^{-x}) = 6$$ $$5e^x - e^{-x} = 6$$

Multiply the entire equation by $e^x$ to eliminate $e^{-x}$:

$$e^x (5e^x - e^{-x}) = 6e^x$$ $$5(e^x)^2 - e^x e^{-x} = 6e^x$$

Since $e^x e^{-x} = e^{x-x} = e^0 = 1$:

$$5(e^x)^2 - 1 = 6e^x$$

Rearrange into a standard quadratic form $ay^2 + by + c = 0$, where $y = e^x$:

$$5(e^x)^2 - 6e^x - 1 = 0$$

Let $y = e^x$. The equation becomes $5y^2 - 6y - 1 = 0$.
Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$$y = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(5)(-1)}}{2(5)}$$ $$y = \frac{6 \pm \sqrt{36 + 20}}{10}$$ $$y = \frac{6 \pm \sqrt{56}}{10}$$

Simplify $\sqrt{56} = \sqrt{4 \times 14} = 2\sqrt{14}$:

$$y = \frac{6 \pm 2\sqrt{14}}{10}$$ $$y = \frac{3 \pm \sqrt{14}}{5}$$

Since $y = e^x$, we have two possible values for $e^x$:

$$e^x = \frac{3 + \sqrt{14}}{5} \quad \text{or} \quad e^x = \frac{3 - \sqrt{14}}{5}$$

For the second case, $\sqrt{14}$ is approximately 3.74. So, $3 - \sqrt{14}$ would be negative. Since $e^x$ must always be positive, the solution $e^x = \frac{3 - \sqrt{14}}{5}$ is not valid.
Therefore, we only consider:

$$e^x = \frac{3 + \sqrt{14}}{5}$$

Take the natural logarithm of both sides:

$$\ln(e^x) = \ln\left(\frac{3 + \sqrt{14}}{5}\right)$$ $$x = \ln\left(\frac{3 + \sqrt{14}}{5}\right)$$

The solution for $x$ is: