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

Take the natural logarithm of both sides: \(\ln(e^x) = \ln(3^{x-2})\).

Using the logarithm power rule, this becomes \(x \ln e = (x-2) \ln 3\).

Since \(\ln e = 1\), the equation simplifies to \(x = (x-2) \ln 3\).

Rearrange to solve for \(x\):

\(x = x \ln 3 - 2 \ln 3\)

\(x - x \ln 3 = -2 \ln 3\)

\(x(1 - \ln 3) = -2 \ln 3\)

\(x = \frac{-2 \ln 3}{1 - \ln 3}\)

Calculate \(x\) using a calculator to get \(x \approx 22.281\).

Start with the equation:

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

Apply the logarithm laws:

\(2 \ln x - \ln 2 = \ln \left( \frac{x^2}{2} \right)\)

So the equation becomes:

\(\ln(2x^2 - 3) = \ln \left( \frac{x^2}{2} \right)\)

Remove the logarithms:

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

Multiply through by 2 to clear the fraction:

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

Rearrange the equation:

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

\(3x^2 = 6\)

Divide by 3:

\(x^2 = 2\)

Take the square root:

\(x = \sqrt{2}\) or \(x = -\sqrt{2}\)

Since \(x\) must be positive (as it is inside a logarithm), the solution is:

\(x = \sqrt{2}\)

Start by dividing both sides of the equation by 2:

\(\ln(5 - e^{-2x}) = \frac{1}{2}\).

Remove the logarithm by exponentiating both sides:

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

Rearrange to solve for \(e^{-2x}\):

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

Take the natural logarithm of both sides:

\(-2x = \ln(5 - e^{\frac{1}{2}})\).

Solve for \(x\):

\(x = -\frac{1}{2} \ln(5 - e^{\frac{1}{2}})\).

Calculate the value to 3 significant figures:

\(x \approx -0.605\).

(i) Start with the given equation:

\(2\ln(4x - 5) + \ln(x + 1) = 3\ln 3\)

Using the properties of logarithms, combine the logs:

\(\ln((4x - 5)^2) + \ln(x + 1) = \ln(27)\)

\(\ln((4x - 5)^2(x + 1)) = \ln(27)\)

Equating the arguments:

\((4x - 5)^2(x + 1) = 27\)

Expanding and simplifying gives:

\(16x^3 - 24x^2 - 15x - 2 = 0\)

(ii) Use the factor theorem to find a root. Testing \(x = 2\):

\(16(2)^3 - 24(2)^2 - 15(2) - 2 = 0\)

So \(x = 2\) is a root, and \((x - 2)\) is a factor.

Divide \(16x^3 - 24x^2 - 15x - 2\) by \((x - 2)\) to find the quotient:

The quotient is \(16x^2 + 8x + 1\).

Factorise further:

\((x - 2)(4x + 1)^2\)

(iii) Solve \(2\ln(4x - 5) + \ln(x + 1) = 3\ln 3\):

From part (i), \((4x - 5)^2(x + 1) = 27\).

Using the factorisation, \(x = 2\) is the only solution.

Start with the equation \(\ln(x+5) = 1 + \ln x\).

We know that \(1 = \ln e\), so the equation becomes \(\ln(x+5) = \ln e + \ln x\).

Using the logarithm property \(\ln a + \ln b = \ln(ab)\), we can rewrite the equation as \(\ln(x+5) = \ln(ex)\).

Since the logarithms are equal, the arguments must be equal: \(x+5 = ex\).

Rearrange to solve for \(x\):

\(x + 5 = ex\)

\(5 = ex - x\)

\(5 = x(e - 1)\)

\(x = \frac{5}{e-1}\)