Start with the equation \(4^{x-2} = 4^x - 4^2\).

Rewrite \(4^{x-2}\) as \(\frac{4^x}{4^2}\), so the equation becomes:

\(\frac{4^x}{16} = 4^x - 16\).

Multiply through by 16 to clear the fraction:

\(4^x = 16(4^x - 16)\).

Expand the right side:

\(4^x = 16 \cdot 4^x - 256\).

Rearrange to isolate terms involving \(4^x\):

\(4^x - 16 \cdot 4^x = -256\).

Factor out \(4^x\):

\(4^x(1 - 16) = -256\).

Simplify:

\(-15 \cdot 4^x = -256\).

Divide both sides by -15:

\(4^x = \frac{256}{15}\).

Take the logarithm of both sides to solve for \(x\):

\(x \log(4) = \log\left(\frac{256}{15}\right)\).

Solve for \(x\):

\(x = \frac{\log\left(\frac{256}{15}\right)}{\log(4)}\).

Calculate the value:

\(x \approx 2.047\).

Start with the equation:

\(\frac{2^x + 1}{2^x - 1} = 5\)

Multiply both sides by \(2^x - 1\):

\(2^x + 1 = 5(2^x - 1)\)

Expand the right side:

\(2^x + 1 = 5 \cdot 2^x - 5\)

Rearrange the terms:

\(2^x + 1 = 5 \cdot 2^x - 5\)

\(1 + 5 = 5 \cdot 2^x - 2^x\)

\(6 = 4 \cdot 2^x\)

Divide both sides by 4:

\(2^x = \frac{6}{4} = 1.5\)

Take the logarithm base 2 of both sides:

\(x = \log_2(1.5)\)

Calculate \(x\) using a calculator:

\(x \approx 0.585\)

Thus, the solution is \(x = 0.585\) to 3 significant figures.

Start with the equation \(3^{x+2} = 3^x + 3^2\).

Rewrite \(3^{x+2}\) as \(3^x imes 3^2\), giving \(3^x imes 9 = 3^x + 9\).

Factor out \(3^x\) from the left side: \(3^x (9 - 1) = 9\).

This simplifies to \(3^x imes 8 = 9\).

Divide both sides by 8: \(3^x = \frac{9}{8}\).

Take the logarithm of both sides: \(x \ln 3 = \ln \left(\frac{9}{8}\right)\).

Solve for \(x\): \(x = \frac{\ln \left(\frac{9}{8}\right)}{\ln 3}\).

Calculate \(x\) to 3 significant figures: \(x \approx 0.107\).

Start by substituting \(u = 3^x\). Then \(3^{-x} = \frac{1}{3^x} = \frac{1}{u}\).

The equation becomes \(u = 2 + \frac{1}{u}\).

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

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

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

Calculate the discriminant: \(b^2 - 4ac = (-2)^2 - 4(1)(-1) = 4 + 4 = 8\).

Thus, \(u = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}\).

Since \(u = 3^x\) must be positive, choose \(u = 1 + \sqrt{2}\).

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

Take the logarithm of both sides: \(x \log 3 = \log(1 + \sqrt{2})\).

Thus, \(x = \frac{\log(1 + \sqrt{2})}{\log 3}\).

Calculate \(x \approx 0.802\) to 3 significant figures.

(i) Given \(y = 2^x\), we have \(2^{-x} = \frac{1}{y}\). Substitute into the equation:

\(2^x - 2^{-x} = y - \frac{1}{y} = 1\).

Multiply through by \(y\) to clear the fraction:

\(y^2 - 1 = y\).

Rearrange to form a quadratic equation:

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

(ii) Solve the quadratic equation \(y^2 - y - 1 = 0\) using the quadratic formula:

\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = -1, c = -1\).

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

Since \(y = 2^x\) and \(y > 0\), choose the positive root:

\(y = \frac{1 + \sqrt{5}}{2}\).

Now solve \(2^x = \frac{1 + \sqrt{5}}{2}\) by taking logarithms:

\(x = \log_2 \left( \frac{1 + \sqrt{5}}{2} \right)\).

Calculate \(x \approx 0.694\).