(i) Given the equation \(2 \ln(4x – 5) + \ln(x + 1) = 3 \ln 3\), use the logarithm laws for product and exponentiation:
Apply the power rule for logarithms:
\(2 \ln(4x – 5) = \ln((4x – 5)^2)\)
\(\ln((4x – 5)^2) + \ln(x + 1) = \ln(27)\)
\(\ln((4x – 5)^2(x + 1)) = \ln(27)\)
Simplify using the property \(\ln a = \ln b \Rightarrow a = b\):
\((4x – 5)^2(x + 1) = 27\)
\(16x^3 – 40x^2 + 25x + 16x^2 – 40x + 25 = 27\)
\(16x^3 – 24x^2 – 15x – 2 = 0\)
(ii) Factorize \(16 x^3 – 24 x^2 – 15 x – 2 = 0\)
Test \(x = 2\) and \(x = -\frac{1}{4}\) as possible roots.
Confirm that \(x = 2\) is a root (as \(16(2)^3 – 24(2)^2 – 15(2) – 2 = 0\)).
Divide the polynomial by \((x – 2)\) using polynomial division:
\[
\begin{array}{r|r}
16x^3 – 24x^2 – 15x – 2 & x – 2 \\
\hline
16x^2 & 16x^3 – 32x^2 \\
8x^2 – 15x & 8x^2 – 16x \\
8x + 1 & 8x^2 – 16x \\
& 8x + 1 \\
& 0 \\
\end{array}
\]
This division gives us \(16x^2 + 8x + 1\) as the quotient. So, the polynomial can be written as:
\( (x – 2)(16x^2 + 8x + 1) = 0\)
(iii) Solve \(2 \ln(4x – 5) + \ln(x + 1) = 3 \ln 3\)
\(2 \ln(4x – 5) = \ln((4x – 5)^2)\)
Using the product rule for logarithms, \(\ln a + \ln b = \ln(ab)\), the equation becomes:
\(\ln((4x – 5)^2) + \ln(x + 1) = \ln((4x – 5)^2(x + 1))\)
Since \(3 \ln 3 = \ln(3^3) = \ln(27)\), we can set the inside of the logarithm equal to 27:
\((4x – 5)^2(x + 1) = 27\)
\(16x^3 – 40x^2 + 25x + 16x^2 – 40x + 25 = 27\)
\(16x^3 – 24x^2 – 15x – 2 = 0\)
We now have a cubic equation to solve: \(16x^3 – 24x^2 – 15x – 2 = 0\).
Solving cubic equations analytically can be complex, but we can start by looking for rational roots using the Rational Root Theorem. Let’s check for simple rational roots such as 1, -1, 2, -2, etc.
After testing, we find that \(x = 2\) is a root (since \(16(2)^3 – 24(2)^2 – 15(2) – 2 = 0\)). We can then divide the cubic polynomial by \((x – 2)\) to find the other roots.
Therefore, the solution to the equation \(2 \ln(4x – 5) + \ln(x + 1) = 3 \ln 3\) is \(x = 2\).