Substitute \(u = 3^x\) into the equation:

\(u + u^2 = u^3\).

Rearrange to form a quadratic equation:

\(u^3 - u^2 - u = 0\).

Factor out \(u\):

\(u(u^2 - u - 1) = 0\).

Since \(u = 3^x\) cannot be zero, solve \(u^2 - u - 1 = 0\).

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

\(u = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}\).

Select the positive root: \(u = \frac{1 + \sqrt{5}}{2}\).

Convert back to \(x\) using \(u = 3^x\):

\(3^x = \frac{1 + \sqrt{5}}{2}\).

Take the logarithm of both sides:

\(x \log 3 = \log \left( \frac{1 + \sqrt{5}}{2} \right)\).

Solve for \(x\):

\(x = \frac{\log \left( \frac{1 + \sqrt{5}}{2} \right)}{\log 3}\).

Calculate \(x\) to 3 significant figures:

\(x \approx 0.438\).

Start with the substitution \(u = 4^x\). The equation becomes:

\(u + 16 = 16u\)

Rearrange to solve for \(u\):

\(16u - u = 16\)

\(15u = 16\)

\(u = \frac{16}{15}\)

Since \(u = 4^x\), we have:

\(4^x = \frac{16}{15}\)

Take the logarithm of both sides:

\(x \log(4) = \log\left(\frac{16}{15}\right)\)

\(x = \frac{\log\left(\frac{16}{15}\right)}{\log(4)}\)

Calculate \(x\) to 3 significant figures:

\(x \approx 0.0466\)

Start with the equation \(5^{x-1} = 5^x - 5\).

Rewrite \(5^{x-1}\) as \(\frac{5^x}{5}\), giving \(\frac{5^x}{5} = 5^x - 5\).

Multiply through by 5 to clear the fraction: \(5^x = 5 \cdot (5^x - 5)\).

This simplifies to \(5^x = 5^{x+1} - 25\).

Rearrange to get \(5^{x+1} - 5^x = 25\).

Factor out \(5^x\): \(5^x(5 - 1) = 25\).

This simplifies to \(4 \cdot 5^x = 25\).

Divide by 4: \(5^x = \frac{25}{4}\).

Take the logarithm of both sides: \(x \ln 5 = \ln \left(\frac{25}{4}\right)\).

Solve for \(x\): \(x = \frac{\ln \left(\frac{25}{4}\right)}{\ln 5}\).

Calculate \(x\) to 3 significant figures: \(x \approx 1.14\).

Start with the equation \(4|5^x - 1| = 5^x\).

This implies two cases: \(4(5^x - 1) = 5^x\) and \(4(1 - 5^x) = 5^x\).

For the first case, \(4(5^x - 1) = 5^x\):

\(4 imes 5^x - 4 = 5^x\)

\(3 imes 5^x = 4\)

\(5^x = \frac{4}{3}\)

Taking logarithms, \(x = \log_5\left(\frac{4}{3}\right)\).

For the second case, \(4(1 - 5^x) = 5^x\):

\(4 - 4 imes 5^x = 5^x\)

\(4 = 5^x + 4 imes 5^x\)

\(4 = 5 imes 5^x\)

\(5^x = \frac{4}{5}\)

Taking logarithms, \(x = \log_5\left(\frac{4}{5}\right)\).

Calculating these values gives \(x = 0.179\) and \(x = -0.139\).

First, consider the equation \(3|2^x - 1| = 2^x\). This can be split into two cases:

Case 1: \(2^x - 1 \geq 0\), so \(|2^x - 1| = 2^x - 1\).

Then the equation becomes \(3(2^x - 1) = 2^x\).

Rearranging gives \(3 \cdot 2^x - 3 = 2^x\).

Thus, \(2^x = \frac{3}{2}\).

Taking logarithms, \(x = \log_2\left(\frac{3}{2}\right)\).

Case 2: \(2^x - 1 < 0\), so \(|2^x - 1| = 1 - 2^x\).

Then the equation becomes \(3(1 - 2^x) = 2^x\).

Rearranging gives \(3 - 3 \cdot 2^x = 2^x\).

Thus, \(2^x = \frac{3}{4}\).

Taking logarithms, \(x = \log_2\left(\frac{3}{4}\right)\).

Calculating these values gives:

\(x = \log_2\left(\frac{3}{2}\right) \approx 0.585\)

\(x = \log_2\left(\frac{3}{4}\right) \approx -0.415\)

Therefore, the solutions are \(x = 0.585\) and \(x = -0.415\).