Past Exam Questions

Start with the equation \(\ln(x^2 + 1) = 1 + 2 \ln x\).

Use the logarithm power rule: \(2 \ln x = \ln(x^2)\).

Substitute to get: \(\ln(x^2 + 1) = 1 + \ln(x^2)\).

Rearrange to: \(\ln(x^2 + 1) - \ln(x^2) = 1\).

Use the logarithm quotient rule: \(\ln\left(\frac{x^2 + 1}{x^2}\right) = 1\).

Exponentiate both sides: \(\frac{x^2 + 1}{x^2} = e\).

Simplify: \(1 + \frac{1}{x^2} = e\).

Rearrange to find \(x^2\): \(\frac{1}{x^2} = e - 1\).

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

Take the square root: \(x = \frac{1}{\sqrt{e - 1}}\).

Calculate \(x\) to 3 significant figures: \(x \approx 0.763\).

To solve the equation \(\ln(1 + 2^x) = 2\), we first remove the logarithm by exponentiating both sides:

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

Next, solve for \(2^x\):

\(2^x = e^2 - 1\).

Now, take the logarithm base 2 of both sides to solve for \(x\):

\(x = \log_2(e^2 - 1)\).

Using a calculator, we find:

\(x \approx 2.676\).

Start with the equation \(\ln(x^2 + 4) = 2 \ln x + \ln 4\).

Apply the logarithm power rule: \(2 \ln x = \ln(x^2)\).

Thus, the equation becomes \(\ln(x^2 + 4) = \ln(x^2) + \ln 4\).

Use the logarithm product rule: \(\ln(x^2) + \ln 4 = \ln(4x^2)\).

Now, the equation is \(\ln(x^2 + 4) = \ln(4x^2)\).

Since the logarithms are equal, set the arguments equal: \(x^2 + 4 = 4x^2\).

Rearrange to get: \(4x^2 - x^2 = 4\).

Simplify: \(3x^2 = 4\).

Divide by 3: \(x^2 = \frac{4}{3}\).

Take the square root: \(x = \frac{2}{\sqrt{3}}\).

Start with the equation \(\ln(x + 4) = 2 \ln x + \ln 4\).

Apply the logarithm power rule: \(2 \ln x = \ln(x^2)\), so the equation becomes \(\ln(x + 4) = \ln(x^2) + \ln 4\).

Use the logarithm product rule: \(\ln(x^2) + \ln 4 = \ln(4x^2)\), so the equation becomes \(\ln(x + 4) = \ln(4x^2)\).

Since the logarithms are equal, set the arguments equal: \(x + 4 = 4x^2\).

Rearrange to form a quadratic equation: \(4x^2 - x - 4 = 0\).

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

Calculate the discriminant: \(b^2 - 4ac = (-1)^2 - 4 \times 4 \times (-4) = 1 + 64 = 65\).

Find the roots: \(x = \frac{-(-1) \pm \sqrt{65}}{8}\).

Calculate: \(x = \frac{1 \pm \sqrt{65}}{8}\).

Only the positive root is valid: \(x = \frac{1 + \sqrt{65}}{8} \approx 1.13\).

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 \(x\):

\(x + 5 = ex\)

\(5 = ex - x\)

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

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

Start with the equation \(\ln(2x + 3) = 2 \ln x + \ln 3\).

Apply the logarithm power rule: \(2 \ln x = \ln(x^2)\).

Substitute: \(\ln(2x + 3) = \ln(x^2) + \ln 3\).

Use the logarithm product rule: \(\ln(x^2) + \ln 3 = \ln(3x^2)\).

Equate the logarithms: \(\ln(2x + 3) = \ln(3x^2)\).

Remove the logarithms: \(2x + 3 = 3x^2\).

Rearrange to form a quadratic equation: \(3x^2 - 2x - 3 = 0\).

Use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -2\), \(c = -3\).

Calculate the discriminant: \(b^2 - 4ac = (-2)^2 - 4 \times 3 \times (-3) = 4 + 36 = 40\).

Find the roots: \(x = \frac{2 \pm \sqrt{40}}{6}\).

Simplify: \(x = \frac{2 \pm 2\sqrt{10}}{6} = \frac{1 \pm \sqrt{10}}{3}\).

Calculate the positive root: \(x = \frac{1 + \sqrt{10}}{3} \approx 1.39\).

Thus, the solution is \(x = 1.39\) to 3 significant figures.

Start with the equation \(\ln(3x + 4) = 2 \ln(x + 1)\).

Apply the logarithm power rule: \(\ln(3x + 4) = \ln((x + 1)^2)\).

Since the logarithms are equal, set the arguments equal: \(3x + 4 = (x + 1)^2\).

Expand the right side: \(3x + 4 = x^2 + 2x + 1\).

Rearrange to form a quadratic equation: \(x^2 - x - 3 = 0\).

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1, b = -1, c = -3\).

Calculate the discriminant: \(b^2 - 4ac = (-1)^2 - 4(1)(-3) = 1 + 12 = 13\).

Find the roots: \(x = \frac{1 \pm \sqrt{13}}{2}\).

Calculate the positive root: \(x = \frac{1 + \sqrt{13}}{2} \approx 2.30\).

Thus, the solution is \(x = 2.30\) to 3 significant figures.

Start with the equation \(\ln(1 + x^2) = 1 + 2 \ln x\).

Use the properties of logarithms: \(\ln a - \ln b = \ln \left(\frac{a}{b}\right)\) and \(\ln a^b = b \ln a\).

Rearrange the equation: \(\ln(1 + x^2) - 2 \ln x = 1\).

This becomes \(\ln \left(\frac{1 + x^2}{x^2}\right) = 1\).

Exponentiate both sides to remove the logarithm: \(\frac{1 + x^2}{x^2} = e^1\).

Simplify: \(1 + x^2 = e x^2\).

Rearrange: \(1 = (e - 1)x^2\).

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

Take the square root: \(x = \sqrt{\frac{1}{e - 1}}\).

Calculate \(x\) using \(e \approx 2.718\):

\(x \approx \sqrt{\frac{1}{2.718 - 1}} \approx 0.763\).

Therefore, the solution is \(x = 0.763\) to 3 significant figures.

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

Use the logarithm property \(\ln a - \ln b = \ln \left( \frac{a}{b} \right)\) to rewrite the equation as:

\(\ln(5-x) = \ln \left( \frac{5}{x} \right)\).

Since the logarithms are equal, their arguments must be equal:

\(5-x = \frac{5}{x}\).

Multiply both sides by \(x\) to eliminate the fraction:

\(x(5-x) = 5\).

Expand and rearrange to form a quadratic equation:

\(x^2 - 5x + 5 = 0\).

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

\(x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}\).

\(x = \frac{5 \pm \sqrt{25 - 20}}{2}\).

\(x = \frac{5 \pm \sqrt{5}}{2}\).

Calculate the roots:

\(x_1 = \frac{5 + \sqrt{5}}{2} \approx 3.62\).

\(x_2 = \frac{5 - \sqrt{5}}{2} \approx 1.38\).

Thus, the solutions are \(x = 1.38\) and \(x = 3.62\).

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

Exponentiate both sides to remove the natural logarithm: \(2 + e^{-x} = e^2\).

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

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

Multiply by -1 to solve for \(x\): \(x = -\ln(e^2 - 2)\).

Calculate the value: \(x \approx -1.68\).

Start with the equation \(\ln(x+2) = 2 + \ln x\).

Use the properties of logarithms to combine the logarithms: \(\ln(x+2) - \ln x = 2\).

This simplifies to \(\ln\left(\frac{x+2}{x}\right) = 2\).

Exponentiate both sides to remove the logarithm: \(\frac{x+2}{x} = e^2\).

Multiply both sides by \(x\) to get \(x+2 = e^2 x\).

Rearrange to solve for \(x\): \(x - e^2 x = -2\).

Factor out \(x\): \(x(1 - e^2) = -2\).

Solve for \(x\): \(x = \frac{-2}{1 - e^2}\).

Calculate \(x\) to three decimal places: \(x \approx 0.313\).

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

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

Rearrange the equation: \(\ln(1 + x) - \ln x = 1\).

This becomes \(\ln \left( \frac{1 + x}{x} \right) = 1\).

Exponentiate both sides to remove the logarithm: \(\frac{1 + x}{x} = e\).

Rearrange to solve for \(x\): \(1 + x = ex\).

\(1 = ex - x\).

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

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

Calculate \(x\) using \(e \approx 2.718\):

\(x \approx \frac{1}{2.718 - 1} \approx 0.582\).

Round to 2 significant figures: \(x = 0.58\).

Start with the equation:

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

Apply the logarithm power rule: \(2 \ln(x + 1) = \ln((x + 1)^2)\)

So the equation becomes:

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

Use the logarithm quotient rule: \(\ln a - \ln b = \ln\left(\frac{a}{b}\right)\)

\(\ln(2x - 1) = \ln\left(\frac{(x + 1)^2}{x}\right)\)

Since the logarithms are equal, the arguments must be equal:

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

Multiply both sides by \(x\) to clear the fraction:

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

Expand both sides:

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

Rearrange to form a quadratic equation:

\(2x^2 - x - x^2 - 2x - 1 = 0\)

\(x^2 - 3x - 1 = 0\)

Solve the quadratic equation using the quadratic formula:

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

Here, \(a = 1\), \(b = -3\), \(c = -1\)

\(x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1}\)

\(x = \frac{3 \pm \sqrt{9 + 4}}{2}\)

\(x = \frac{3 \pm \sqrt{13}}{2}\)

Calculate the positive root:

\(x = \frac{3 + \sqrt{13}}{2} \approx 3.303\)

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

\(\ln(2^{3x-1}) = \ln(5 \cdot 3^{1-x})\)

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

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

Expand and rearrange the equation:

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

Combine like terms:

\(3x \ln 2 + x \ln 3 = \ln 5 + \ln 3 + \ln 2\)

Factor out \(x\):

\(x(3 \ln 2 + \ln 3) = \ln 5 + \ln 3 + \ln 2\)

Solve for \(x\):

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

Combine the logarithms:

\(x = \frac{\ln(5 \cdot 3 \cdot 2)}{\ln(2^3 \cdot 3)}\)

\(x = \frac{\ln 30}{\ln 24}\)

Start with the equation \(2^{3x-1} = 5 \cdot 3^{-x}\).

Take the natural logarithm of both sides: \(\ln(2^{3x-1}) = \ln(5 \cdot 3^{-x})\).

Apply the logarithm laws: \((3x-1)\ln 2 = \ln 5 + (-x)\ln 3\).

Rearrange to form a linear equation: \(3x \ln 2 - x \ln 3 = \ln 5 + \ln 2\).

Factor out \(x\): \(x(3 \ln 2 - \ln 3) = \ln 5 + \ln 2\).

Solve for \(x\): \(x = \frac{\ln 5 + \ln 2}{3 \ln 2 - \ln 3}\).

Simplify the expression: \(x = \frac{\ln 10}{\ln 24}\).

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

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

Rearrange the equation: \(\ln(e^{2x} + 3) - \ln 3 = 2x\).

This becomes \(\ln\left(\frac{e^{2x} + 3}{3}\right) = 2x\).

Exponentiate both sides to remove the logarithm: \(\frac{e^{2x} + 3}{3} = e^{2x}\).

Multiply through by 3: \(e^{2x} + 3 = 3e^{2x}\).

Rearrange to form a quadratic in terms of \(e^{2x}\): \(3 = 2e^{2x}\).

Divide by 2: \(e^{2x} = \frac{3}{2}\).

Take the natural logarithm of both sides: \(2x = \ln\left(\frac{3}{2}\right)\).

Solve for \(x\): \(x = \frac{1}{2} \ln\left(\frac{3}{2}\right)\).

Calculate \(x\) to 3 decimal places: \(x \approx 0.203\).