Let \(u = 4^x\). Then \(4^{-x} = \frac{1}{u}\).

The equation becomes \(u = 3 + \frac{1}{u}\).

Multiply through by \(u\) to clear the fraction: \(u^2 = 3u + 1\).

Rearrange to form a quadratic equation: \(u^2 - 3u - 1 = 0\).

Use the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1\), \(b = -3\), \(c = -1\).

\(u = \frac{3 \pm \sqrt{9 + 4}}{2} = \frac{3 \pm \sqrt{13}}{2}\).

Since \(u = 4^x\) must be positive, take the positive root: \(u = \frac{3 + \sqrt{13}}{2}\).

Calculate \(u \approx 3.30277563773\).

Since \(u = 4^x\), solve for \(x\): \(x = \log_4(u)\).

\(x = \frac{\log(u)}{\log(4)}\).

\(x \approx \frac{\log(3.30277563773)}{\log(4)} \approx 0.862\).

Let \(y = 3^x\). Then \(3^{-x} = \frac{1}{y}\).

The equation becomes \(\frac{y^2 + \frac{1}{y}}{y^2 - \frac{1}{y}} = 4\).

Multiply numerator and denominator by \(y\) to clear the fraction:

\(\frac{y^3 + 1}{y^3 - 1} = 4\).

Cross-multiply to get:

\(y^3 + 1 = 4(y^3 - 1)\).

Expand and simplify:

\(y^3 + 1 = 4y^3 - 4\).

Rearrange terms:

\(3y^3 = 5\).

So, \(y^3 = \frac{5}{3}\).

Since \(y = 3^x\), we have \((3^x)^3 = \frac{5}{3}\).

Thus, \(3^{3x} = \frac{5}{3}\).

Take the logarithm of both sides:

\(3x \log 3 = \log \left( \frac{5}{3} \right)\).

Solve for \(x\):

\(x = \frac{\log \left( \frac{5}{3} \right)}{3 \log 3}\).

Calculate \(x\) to 3 decimal places:

\(x \approx 0.155\).

First, express \(9^x\) in terms of base 3: \(9^x = (3^2)^x = (3^x)^2\).

Let \(u = 3^x\). Then the equation becomes \(u^2 = u + 12\).

Rearrange to form a quadratic equation: \(u^2 - u - 12 = 0\).

Factor the quadratic: \((u - 4)(u + 3) = 0\).

Thus, \(u = 4\) or \(u = -3\). Since \(u = 3^x\) and \(3^x > 0\), we have \(u = 4\).

Therefore, \(3^x = 4\).

Take the natural logarithm of both sides: \(x \ln 3 = \ln 4\).

Solve for \(x\): \(x = \frac{\ln 4}{\ln 3} \approx 1.26186\).

Round to two decimal places: \(x = 1.26\).

Let \(u = 5^x\). Then the equation becomes \(u^2 = u + 5\).

Rearrange to form a quadratic equation: \(u^2 - u - 5 = 0\).

Use the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -1\), \(c = -5\).

Calculate the discriminant: \(b^2 - 4ac = (-1)^2 - 4 \times 1 \times (-5) = 1 + 20 = 21\).

Thus, \(u = \frac{1 \pm \sqrt{21}}{2}\).

Since \(u = 5^x\) must be positive, choose the positive root: \(u = \frac{1 + \sqrt{21}}{2}\).

Now, solve for \(x\): \(5^x = \frac{1 + \sqrt{21}}{2}\).

Take the logarithm of both sides: \(x \log 5 = \log \left(\frac{1 + \sqrt{21}}{2}\right)\).

Thus, \(x = \frac{\log \left(\frac{1 + \sqrt{21}}{2}\right)}{\log 5}\).

Calculate \(x\) to 3 decimal places: \(x \approx 0.638\).

Start with the equation:

\(\frac{3^x + 2}{3^x - 2} = 8\)

Cross-multiply to eliminate the fraction:

\(3^x + 2 = 8(3^x - 2)\)

Expand the right side:

\(3^x + 2 = 8 \times 3^x - 16\)

Rearrange to isolate terms involving \(3^x\):

\(3^x + 2 = 8 \times 3^x - 16\)

\(2 + 16 = 8 \times 3^x - 3^x\)

\(18 = 7 \times 3^x\)

Divide both sides by 7:

\(3^x = \frac{18}{7}\)

Take the logarithm of both sides:

\(x \log 3 = \log \left( \frac{18}{7} \right)\)

Solve for \(x\):

\(x = \frac{\log \left( \frac{18}{7} \right)}{\log 3}\)

Calculate \(x\) to 3 decimal places:

\(x \approx 0.860\)