(a) Start with the equation \(\log_3(2x + 1) = 1 + 2\log_3(x - 1)\).

Using the law of logarithms, \(2\log_3(x - 1) = \log_3((x - 1)^2)\).

Thus, the equation becomes \(\log_3(2x + 1) = \log_3(3) + \log_3((x - 1)^2)\).

This simplifies to \(\log_3(2x + 1) = \log_3(3(x - 1)^2)\).

Equating the arguments gives \(2x + 1 = 3(x - 1)^2\).

Expanding \(3(x - 1)^2\) gives \(3(x^2 - 2x + 1) = 3x^2 - 6x + 3\).

Thus, \(2x + 1 = 3x^2 - 6x + 3\).

Rearranging gives \(3x^2 - 8x + 2 = 0\).

(b) Use the quadratic equation from part (a) to solve for \(y\):

\(3(2y)^2 - 8(2y) + 2 = 0\) simplifies to \(12y^2 - 16y + 2 = 0\).

Divide through by 2: \(6y^2 - 8y + 1 = 0\).

Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 6, b = -8, c = 1\).

\(y = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 6 \cdot 1}}{12}\).

\(y = \frac{8 \pm \sqrt{64 - 24}}{12}\).

\(y = \frac{8 \pm \sqrt{40}}{12}\).

\(y = \frac{8 \pm 2\sqrt{10}}{12}\).

\(y = \frac{4 \pm \sqrt{10}}{6}\).

Calculating gives \(y \approx 1.19\) (to 2 decimal places).

Start by using the property \(\log_{10} 10 = 1\) and \(\log_{10} 10^{-1} = -1\).

Rearrange the equation:

\(\log_{10}(2x + 1) = 2\log_{10}(x + 1) - 1\)

\(\log_{10}(2x + 1) + 1 = 2\log_{10}(x + 1)\)

\(\log_{10}(10(2x + 1)) = \log_{10}((x + 1)^2)\)

Equate the arguments:

\(10(2x + 1) = (x + 1)^2\)

Expand and simplify:

\(20x + 10 = x^2 + 2x + 1\)

\(x^2 - 18x - 9 = 0\)

Solve the quadratic equation using the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where \(a = 1, b = -18, c = -9\).

\(x = \frac{18 \pm \sqrt{324 + 36}}{2}\)

\(x = \frac{18 \pm \sqrt{360}}{2}\)

\(x = \frac{18 \pm 18.9737}{2}\)

\(x = 18.487\) and \(x = -0.487\)

(i) Start with the equation \(\log_{10}(x-4) = 2 - \log_{10} x\).

Using the law of logarithms for subtraction, \(\log_{10}(x-4) + \log_{10} x = 2\).

This can be rewritten as \(\log_{10}((x-4)x) = 2\).

Since \(\log_{10} 100 = 2\), we have \((x-4)x = 100\).

Expanding gives \(x^2 - 4x = 100\).

Rearranging, we obtain the quadratic equation \(x^2 - 4x - 100 = 0\).

(ii) Solve the quadratic equation \(x^2 - 4x - 100 = 0\).

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -4\), and \(c = -100\).

Calculate the discriminant: \(b^2 - 4ac = (-4)^2 - 4 \times 1 \times (-100) = 16 + 400 = 416\).

Thus, \(x = \frac{4 \pm \sqrt{416}}{2}\).

Calculate \(\sqrt{416} \approx 20.396\).

So, \(x = \frac{4 \pm 20.396}{2}\).

This gives solutions \(x = 12.198\) and \(x = -8.198\).

Since \(x\) must be positive, the solution is \(x = 12.2\) (to 3 significant figures).

Start with the equation: \(2\log_2 x = 3 + \log_2(x + 1)\).

Apply the power rule of logarithms: \(\log_2 x^2 = 3 + \log_2(x + 1)\).

Use the property \(\log_a b - \log_a c = \log_a \left(\frac{b}{c}\right)\):

\(\log_2 \left(\frac{x^2}{x+1}\right) = 3\).

Convert the logarithmic equation to an exponential equation:

\(\frac{x^2}{x+1} = 2^3 = 8\).

Multiply both sides by \(x+1\):

\(x^2 = 8(x + 1)\).

Expand and rearrange to form a quadratic equation:

\(x^2 - 8x - 8 = 0\).

Solve the quadratic equation using the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -8\), \(c = -8\).

Calculate the discriminant: \(b^2 - 4ac = (-8)^2 - 4(1)(-8) = 64 + 32 = 96\).

\(x = \frac{8 \pm \sqrt{96}}{2}\).

\(x = \frac{8 \pm 4\sqrt{6}}{2}\).

\(x = 4 \pm 2\sqrt{6}\).

Calculate the positive root: \(x = 4 + 2\sqrt{6} \approx 8.90\).

Thus, the solution is \(x = 8.90\).

Start with the equation \(\log_{10}(x+9) = 2 + \log_{10} x\).

Use the logarithm property: \(\log_{10}(x+9) = \log_{10}(10^2) + \log_{10} x\).

This simplifies to \(\log_{10}(x+9) = \log_{10}(100x)\).

Since the logarithms are equal, set the arguments equal: \(x+9 = 100x\).

Rearrange to solve for \(x\):

\(x + 9 = 100x\)

\(9 = 99x\)

\(x = \frac{9}{99} = \frac{1}{11}\).

(i) Start with the equation \(\log_2(x+5) = 5 - \log_2 x\).

Using the law of logarithms, \(\log_2(x+5) + \log_2 x = 5\).

This can be rewritten as \(\log_2((x+5)x) = 5\).

Therefore, \((x+5)x = 2^5\).

So, \(x^2 + 5x = 32\).

Rearranging gives the quadratic equation \(x^2 + 5x - 32 = 0\).

(ii) Solve the quadratic equation \(x^2 + 5x - 32 = 0\).

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 5\), \(c = -32\).

Calculate the discriminant: \(b^2 - 4ac = 5^2 - 4 \times 1 \times (-32) = 25 + 128 = 153\).

Thus, \(x = \frac{-5 \pm \sqrt{153}}{2}\).

The positive solution is \(x = \frac{\sqrt{153} - 5}{2}\), which is approximately \(x = 3.68\).

(i) Start with the equation \(\log_{10}(x+5) = 2 - \log_{10} x\).

Use the property of logarithms: \(\log_{10} a - \log_{10} b = \log_{10} \left( \frac{a}{b} \right)\).

Rearrange the equation: \(\log_{10}(x+5) + \log_{10} x = 2\).

Combine the logarithms: \(\log_{10}((x+5)x) = 2\).

Convert the logarithmic equation to an exponential equation: \((x+5)x = 10^2\).

Simplify: \(x^2 + 5x = 100\).

Thus, the quadratic equation is \(x^2 + 5x - 100 = 0\).

(ii) Solve the quadratic equation \(x^2 + 5x - 100 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 5\), and \(c = -100\).

Calculate the discriminant: \(b^2 - 4ac = 5^2 - 4 \times 1 \times (-100) = 25 + 400 = 425\).

Find the roots: \(x = \frac{-5 \pm \sqrt{425}}{2}\).

Since \(x\) must be positive, choose the positive root: \(x = \frac{\sqrt{425} - 5}{2}\).

Approximate the value: \(x \approx 7.81\).