Past Exam Questions

Start with the equation \(4|5^x - 1| = 5^x\).

This implies two cases: \(4(5^x - 1) = 5^x\) and \(4(1 - 5^x) = 5^x\).

For the first case, \(4(5^x - 1) = 5^x\):

\(4 \times 5^x - 4 = 5^x\)

\(3 \times 5^x = 4\)

\(5^x = \frac{4}{3}\)

Taking logarithms, \(x = \log_5\left(\frac{4}{3}\right)\).

For the second case, \(4(1 - 5^x) = 5^x\):

\(4 - 4 \times 5^x = 5^x\)

\(4 = 5^x + 4 \times 5^x\)

\(4 = 5 \times 5^x\)

\(5^x = \frac{4}{5}\)

Taking logarithms, \(x = \log_5\left(\frac{4}{5}\right)\).

Calculating these values gives \(x = 0.179\) and \(x = -0.139\).

First, consider the equation \(3|2^x - 1| = 2^x\). This can be split into two cases:

Case 1: \(2^x - 1 \geq 0\), so \(|2^x - 1| = 2^x - 1\).

Then the equation becomes \(3(2^x - 1) = 2^x\).

Rearranging gives \(3 \cdot 2^x - 3 = 2^x\).

Thus, \(2^x = \frac{3}{2}\).

Taking logarithms, \(x = \log_2\left(\frac{3}{2}\right)\).

Case 2: \(2^x - 1 < 0\), so \(|2^x - 1| = 1 - 2^x\).

Then the equation becomes \(3(1 - 2^x) = 2^x\).

Rearranging gives \(3 - 3 \cdot 2^x = 2^x\).

Thus, \(2^x = \frac{3}{4}\).

Taking logarithms, \(x = \log_2\left(\frac{3}{4}\right)\).

Calculating these values gives:

\(x = \log_2\left(\frac{3}{2}\right) \approx 0.585\)

\(x = \log_2\left(\frac{3}{4}\right) \approx -0.415\)

Therefore, the solutions are \(x = 0.585\) and \(x = -0.415\).

(i) Consider the equation \(2|x - 1| = 3|x|\). This can be split into two cases:

Case 1: \(2(x - 1) = 3x\)

\(2x - 2 = 3x\)

\(-2 = x\)

Case 2: \(2(1 - x) = 3x\)

\(2 - 2x = 3x\)

\(2 = 5x\)

\(x = \frac{2}{5}\)

Thus, the solutions are \(x = -2\) and \(x = \frac{2}{5}\).

(ii) For the equation \(2|5^x - 1| = 3|5^x|\), we use the same approach:

Case 1: \(2(5^x - 1) = 3(5^x)\)

\(2 \cdot 5^x - 2 = 3 \cdot 5^x\)

\(-2 = 5^x\)

Case 2: \(2(1 - 5^x) = 3(5^x)\)

\(2 - 2 \cdot 5^x = 3 \cdot 5^x\)

\(2 = 5 \cdot 5^x\)

\(5^x = \frac{2}{5}\)

Taking logarithms, \(x = \log_5\left(\frac{2}{5}\right)\)

Using a calculator, \(x \approx 0.569\) to 3 significant figures.

We start with the equation \(2|3^x - 1| = 3^x\).

This can be split into two cases:

Case 1: \(2(3^x - 1) = 3^x\)

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

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

\(3^x = 2\)

Taking logarithms, \(x = \log_3 2\).

Calculating, \(x \approx 0.631\).

Case 2: \(2(1 - 3^x) = 3^x\)

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

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

\(2 = 3 \cdot 3^x\)

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

Taking logarithms, \(x = \log_3 \frac{2}{3}\).

Calculating, \(x \approx -0.369\).

Thus, the solutions are \(x \approx 0.631\) and \(x \approx -0.369\).

(i) To solve \(|4x - 1| = |x - 3|\), consider the cases:

Case 1: \(4x - 1 = x - 3\)

\(4x - x = -3 + 1\)

\(3x = -2\)

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

Case 2: \(4x - 1 = -(x - 3)\)

\(4x - 1 = -x + 3\)

\(4x + x = 3 + 1\)

\(5x = 4\)

\(x = \frac{4}{5}\)

Thus, the solutions are \(x = -\frac{2}{3}\) and \(x = \frac{4}{5}\).

(ii) Using the result from part (i), we have \(4^y = \frac{4}{5}\).

Taking logarithms, \(y \log 4 = \log \frac{4}{5}\).

\(y = \frac{\log \frac{4}{5}}{\log 4}\).

Calculating gives \(y \approx -0.161\).

The equation given is \(|4 - 2^x| = 10\). This implies two cases:

1. \(4 - 2^x = 10\)

2. \(4 - 2^x = -10\)

For the first case:

\(4 - 2^x = 10\)

\(-2^x = 10 - 4\)

\(-2^x = 6\)

This case is not possible as \(2^x\) is always positive.

For the second case:

\(4 - 2^x = -10\)

\(-2^x = -10 - 4\)

\(-2^x = -14\)

\(2^x = 14\)

Taking logarithms on both sides:

\(x \log 2 = \log 14\)

\(x = \frac{\log 14}{\log 2}\)

Calculating the value:

\(x \approx 3.81\)

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

\(\ln(5^{3-2x}) = \ln(4 \cdot 7^x)\)

Apply the logarithm of a product rule: \(\ln(a \cdot b) = \ln a + \ln b\):

\(\ln(5^{3-2x}) = \ln 4 + \ln(7^x)\)

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

\((3-2x) \ln 5 = \ln 4 + x \ln 7\)

Expand and rearrange to form a linear equation in \(x\):

\(3 \ln 5 - 2x \ln 5 = \ln 4 + x \ln 7\)

Combine like terms:

\(3 \ln 5 - \ln 4 = x \ln 7 + 2x \ln 5\)

Factor out \(x\):

\(3 \ln 5 - \ln 4 = x(\ln 7 + 2 \ln 5)\)

Solve for \(x\):

\(x = \frac{3 \ln 5 - \ln 4}{\ln 7 + 2 \ln 5}\)

Calculate the value of \(x\) to three decimal places:

\(x \approx 0.666\)

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

\(\ln(4^{3x-1}) = \ln(3 \cdot 5^x)\)

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

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

Expand and rearrange the equation to solve for \(x\):

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

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

Factor out \(x\):

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

Solve for \(x\):

\(x = \frac{\ln 3 + \ln 4}{3 \ln 4 - \ln 5}\)

Calculate the value of \(x\) to three decimal places:

\(x \approx 0.975\)

Take the natural logarithm of both sides of the equation:

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

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

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

Expand the right side:

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

Rearrange to solve for \(x\):

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

Factor out \(x\):

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

Solve for \(x\):

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

Calculate \(x\) to 3 significant figures:

\(x \approx 0.866\)

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

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

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

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

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

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

Calculate \(x\) to 3 significant figures: \(x = 1.09\).

First, consider the inequality \(|2^{x+1} - 2| < 0.5\). This can be rewritten as two separate inequalities:

1. \(2^{x+1} - 2 < 0.5\)

2. \(2^{x+1} - 2 > -0.5\)

For the first inequality:

\(2^{x+1} < 2.5\)

Taking the logarithm base 2, we have:

\((x+1)\ln 2 < \ln 2.5\)

\(x < \frac{\ln 2.5}{\ln 2} - 1\)

For the second inequality:

\(2^{x+1} > 1.5\)

Taking the logarithm base 2, we have:

\((x+1)\ln 2 > \ln 1.5\)

\(x > \frac{\ln 1.5}{\ln 2} - 1\)

Calculating the critical values:

\(x < \frac{\ln 2.5}{\ln 2} - 1 \approx 0.322\)

\(x > \frac{\ln 1.5}{\ln 2} - 1 \approx -0.415\)

Thus, the solution is \(-0.415 < x < 0.322\).

Start with the inequality \(2(3^{1-2x}) < 5^x\).

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

Apply the logarithm power rule: \(\ln(2) + (1-2x)\ln(3) < x \ln(5)\).

Rearrange the inequality: \(\ln(2) + \ln(3) - 2x \ln(3) < x \ln(5)\).

Combine like terms: \(\ln(6) < x(\ln(5) + 2\ln(3))\).

Solve for \(x\): \(x > \frac{\ln(6)}{\ln(5) + 2\ln(3)}\).

Simplify the denominator: \(x > \frac{\ln(6)}{\ln(45)}\).

To solve the inequality \(|3^x - 8| < 0.5\), we consider the two cases:

1. \(3^x - 8 < 0.5\)

2. \(3^x - 8 > -0.5\)

For the first case:

\(3^x < 8.5\)

Taking logarithms, \(x < \log_3(8.5)\)

For the second case:

\(3^x > 7.5\)

Taking logarithms, \(x > \log_3(7.5)\)

Calculating the logarithms to 3 significant figures:

\(\log_3(8.5) \approx 1.95\)

\(\log_3(7.5) \approx 1.83\)

Thus, the solution is \(1.83 < x < 1.95\).

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:

\(x \approx 1.52\)