(i) Consider the equation \(2|x - 1| = 3|x|\). This can be split into two cases:

Case 1: \(2(x - 1) = 3x\)

\(2x - 2 = 3x\)

\(-2 = x\)

Case 2: \(2(1 - x) = 3x\)

\(2 - 2x = 3x\)

\(2 = 5x\)

\(x = \frac{2}{5}\)

Thus, the solutions are \(x = -2\) and \(x = \frac{2}{5}\).

(ii) For the equation \(2|5^x - 1| = 3|5^x|\), we use the same approach:

Case 1: \(2(5^x - 1) = 3(5^x)\)

\(2 \cdot 5^x - 2 = 3 \cdot 5^x\)

\(-2 = 5^x\)

Case 2: \(2(1 - 5^x) = 3(5^x)\)

\(2 - 2 \cdot 5^x = 3 \cdot 5^x\)

\(2 = 5 \cdot 5^x\)

\(5^x = \frac{2}{5}\)

Taking logarithms, \(x = \log_5\left(\frac{2}{5}\right)\)

Using a calculator, \(x \approx 0.569\) to 3 significant figures.

We start with the equation \(2|3^x - 1| = 3^x\).

This can be split into two cases:

Case 1: \(2(3^x - 1) = 3^x\)

\(2 \cdot 3^x - 2 = 3^x\)

\(2 \cdot 3^x - 3^x = 2\)

\(3^x = 2\)

Taking logarithms, \(x = \log_3 2\).

Calculating, \(x \approx 0.631\).

Case 2: \(2(1 - 3^x) = 3^x\)

\(2 - 2 \cdot 3^x = 3^x\)

\(2 = 3^x + 2 \cdot 3^x\)

\(2 = 3 \cdot 3^x\)

\(3^x = \frac{2}{3}\)

Taking logarithms, \(x = \log_3 \frac{2}{3}\).

Calculating, \(x \approx -0.369\).

Thus, the solutions are \(x \approx 0.631\) and \(x \approx -0.369\).

(i) To solve \(|4x - 1| = |x - 3|\), consider the cases:

Case 1: \(4x - 1 = x - 3\)

\(4x - x = -3 + 1\)

\(3x = -2\)

\(x = -\frac{2}{3}\)

Case 2: \(4x - 1 = -(x - 3)\)

\(4x - 1 = -x + 3\)

\(4x + x = 3 + 1\)

\(5x = 4\)

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

Thus, the solutions are \(x = -\frac{2}{3}\) and \(x = \frac{4}{5}\).

(ii) Using the result from part (i), we have \(4^y = \frac{4}{5}\).

Taking logarithms, \(y \log 4 = \log \frac{4}{5}\).

\(y = \frac{\log \frac{4}{5}}{\log 4}\).

Calculating gives \(y \approx -0.161\).

The equation given is \(|4 - 2^x| = 10\). This implies two cases:

1. \(4 - 2^x = 10\)

2. \(4 - 2^x = -10\)

For the first case:

\(4 - 2^x = 10\)

\(-2^x = 10 - 4\)

\(-2^x = 6\)

This case is not possible as \(2^x\) is always positive.

For the second case:

\(4 - 2^x = -10\)

\(-2^x = -10 - 4\)

\(-2^x = -14\)

\(2^x = 14\)

Taking logarithms on both sides:

\(x \log 2 = \log 14\)

\(x = \frac{\log 14}{\log 2}\)

Calculating the value:

\(x \approx 3.81\)

Start by taking the natural logarithm of both sides of the equation:

\(\ln(5^{3-2x}) = \ln(4 \cdot 7^x)\)

Apply the logarithm of a product rule: \(\ln(a \cdot b) = \ln a + \ln b\):

\(\ln(5^{3-2x}) = \ln 4 + \ln(7^x)\)

Apply the power rule: \(\ln(a^b) = b \ln a\):

\((3-2x) \ln 5 = \ln 4 + x \ln 7\)

Expand and rearrange to form a linear equation in \(x\):

\(3 \ln 5 - 2x \ln 5 = \ln 4 + x \ln 7\)

Combine like terms:

\(3 \ln 5 - \ln 4 = x \ln 7 + 2x \ln 5\)

Factor out \(x\):

\(3 \ln 5 - \ln 4 = x(\ln 7 + 2 \ln 5)\)

Solve for \(x\):

\(x = \frac{3 \ln 5 - \ln 4}{\ln 7 + 2 \ln 5}\)

Calculate the value of \(x\) to three decimal places:

\(x \approx 0.666\)