Solve the equation $e^{x-2} - 14\left(\frac{1}{e}\right)^{x-2} = 5$. Solve the equation $log_{2}(3x+1) - \dfrac{2}{\log_{\sqrt{2}}2} = \log_{2}(2 - x^{2})$.

1. Introduce a positive substitution to simplify the exponential terms.

Let $y=e^{x-2}$. Since $e^{x-2}>0$, $y>0$.

Also,
${\left(\frac{1}{e}\right)}^{x-2}=e^{-(x-2)}=\frac{1}{e^{x-2}}=\frac{1}{y}.$

2. Form and solve the resulting quadratic equation.

The equation becomes
$y-14\left(\frac{1}{y}\right)=5.$

Multiply throughout by $y$:
$y^2-14=5y$
$y^2-5y-14=0.$

Factorise:
$(y-7)(y+2)=0.$

So $y=7$ or $y=-2$.

3. Reject the invalid substituted value and solve for $x$.

Since $y=e^{x-2}>0$, reject $y=-2$.

Thus
$e^{x-2}=7.$

Taking natural logarithms,
$x-2=\ln 7$
$x=2+\ln 7.$

1. Simplify the constant involving the logarithm with base $\sqrt{2}$.

Since $\sqrt{2}=2^{\frac12}$,
$\log_{\sqrt{2}}2=\log_{2^{\frac12}}2=2.$
Hence
$\frac{2}{\log_{\sqrt{2}}2}=\frac{2}{2}=1.$

2. Combine the logarithmic terms into one logarithm equation.

The equation becomes
$\log_2(3x+1)-1=\log_2(2-x^2).$
Since $1=\log_2 2$,
$\log_2(3x+1)-\log_2 2=\log_2(2-x^2),$
$\log_2\left(\frac{3x+1}{2}\right)=\log_2(2-x^2).$

3. Equate the arguments and solve the resulting quadratic equation.

Since the logarithms have the same base,
$\frac{3x+1}{2}=2-x^2.$
$3x+1=4-2x^2$
$2x^2+3x-3=0.$
Using the quadratic formula,
$x=\frac{-3\pm\sqrt{3^2-4(2)(-3)}}{2(2)}$
$x=\frac{-3\pm\sqrt{33}}{4}.$

4. Check the domain restrictions of the logarithms and reject invalid solution.

For the logarithms to be defined,
$3x+1>0 \quad \text{and} \quad 2-x^2>0.$
So
$x>-\frac13 \quad \text{and} \quad -\sqrt2<x<\sqrt2.$
Now
$\frac{-3+\sqrt{33}}{4}\approx0.686,$
which is valid, but
$\frac{-3-\sqrt{33}}{4}\approx-2.186,$
which is invalid.
Therefore,
$x=\frac{-3+\sqrt{33}}{4}.$