Past Exam Questions

To solve the inequality \(|2^x - 8| < 5\), we consider two cases:

1. \(2^x - 8 < 5\)

2. \(2^x - 8 > -5\)

For the first case, \(2^x - 8 < 5\):

\(2^x < 13\)

Taking logarithm base 2 on both sides, we get:

\(x < \log_2 13\)

For the second case, \(2^x - 8 > -5\):

\(2^x > 3\)

Taking logarithm base 2 on both sides, we get:

\(x > \log_2 3\)

Thus, combining both inequalities, we have:

\(\log_2 3 < x < \log_2 13\)

Calculating the logarithms, we find:

\(\log_2 3 \approx 1.58\) and \(\log_2 13 \approx 3.70\)

Therefore, the solution is \(1.58 < x < 3.70\).

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

Subtract \(\ln x\) from both sides to get \(\ln(x+5) - \ln x = 5\).

Use the logarithmic identity \(\ln a - \ln b = \ln \left( \frac{a}{b} \right)\) to rewrite the equation as \(\ln \left( \frac{x+5}{x} \right) = 5\).

Exponentiate both sides to remove the logarithm: \(\frac{x+5}{x} = e^5\).

Multiply both sides by \(x\) to get \(x+5 = xe^5\).

Rearrange to solve for \(x\): \(x - xe^5 = -5\).

Factor out \(x\): \(x(1 - e^5) = -5\).

Solve for \(x\): \(x = \frac{-5}{1 - e^5}\).

Calculate the value: \(x \approx 0.034\).

Start with the equation:

\(\ln(1 + e^{-3x}) = 2\)

Exponentiate both sides to remove the natural logarithm:

\(1 + e^{-3x} = e^2\)

Rearrange to solve for \(e^{-3x}\):

\(e^{-3x} = e^2 - 1\)

Calculate \(e^2 - 1\):

\(e^2 \approx 7.389\), so \(e^2 - 1 \approx 6.389\)

Now solve for \(x\):

\(e^{-3x} = 6.389\)

Take the natural logarithm of both sides:

\(-3x = \ln(6.389)\)

Calculate \(\ln(6.389) \approx 1.855\)

So:

\(-3x = 1.855\)

Divide by -3:

\(x = -\frac{1.855}{3} \approx -0.618\)

Thus, the solution is \(x = -0.618\) to 3 decimal places.

Start by using the properties of logarithms to combine and simplify the equation:

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

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

Equating the two expressions gives:

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

Remove the logarithms by setting the arguments equal:

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

Expand both sides:

\(6x + 15 = x^2 + 4x + 4\)

Rearrange to form a quadratic equation:

\(x^2 - 2x - 11 = 0\)

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1, b = -2, c = -11\):

\(x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-11)}}{2 \cdot 1}\)

\(x = \frac{2 \pm \sqrt{4 + 44}}{2}\)

\(x = \frac{2 \pm \sqrt{48}}{2}\)

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

\(x = 1 \pm 2\sqrt{3}\)

Thus, the solutions are \(x = 1 + 2\sqrt{3}\) or \(x = 1 - 2\sqrt{3}\). Since \(x = 1 - 2\sqrt{3}\) is not valid in the original logarithmic equation (as it results in a negative argument), the valid solution is \(x = 1 + 2\sqrt{3}\).

Start by dividing both sides of the equation by 5:

\(\ln(4 - 3^x) = \frac{6}{5}\).

Exponentiate both sides to remove the logarithm:

\(4 - 3^x = e^{\frac{6}{5}}\).

Calculate \(e^{\frac{6}{5}}\) to get approximately 3.3201169.

Thus, \(4 - 3^x = 3.3201169\).

Rearrange to solve for \(3^x\):

\(3^x = 4 - 3.3201169\).

\(3^x = 0.6798831\).

Take the logarithm base 3 of both sides:

\(x = \log_3(0.6798831)\).

Calculate \(x\) to get approximately \(-0.351\).

Start by using the properties of logarithms to combine the terms on the right-hand side:

\(\ln(2x - 3) = \ln\left(\frac{x^2}{x-1}\right)\).

Since the logarithms are equal, we equate the arguments:

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

Multiply both sides by \(x-1\) to clear the fraction:

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

Expand the left-hand side:

\(2x^2 - 2x - 3x + 3 = x^2\).

Simplify and combine like terms:

\(2x^2 - 5x + 3 = x^2\).

Subtract \(x^2\) from both sides:

\(x^2 - 5x + 3 = 0\).

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1, b = -5, c = 3\):

\(x = \frac{5 \pm \sqrt{25 - 12}}{2}\).

\(x = \frac{5 \pm \sqrt{13}}{2}\).

Calculate the roots:

\(x = \frac{5 + \sqrt{13}}{2} \approx 4.30\) and \(x = \frac{5 - \sqrt{13}}{2} \approx 0.70\).

Since \(x\) must be greater than 1 for the logarithms to be defined, the solution is:

\(x = 4.30\).

Start with the equation:

\(\ln(x^4 - 4) = 4 \ln x - \ln 4\)

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

\(\ln(x^4 - 4) = \ln \left( \frac{x^4}{4} \right)\)

Since the logarithms are equal, set the arguments equal:

\(x^4 - 4 = \frac{x^4}{4}\)

Multiply through by 4 to clear the fraction:

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

Expand and simplify:

\(4x^4 - 16 = x^4\)

Rearrange to form a quadratic equation:

\(3x^4 = 16\)

Divide by 3:

\(x^4 = \frac{16}{3}\)

Take the fourth root of both sides:

\(x = \left( \frac{16}{3} \right)^{1/4}\)

Calculate the value:

\(x \approx 1.52\)

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 \(x = 3.62\).