Past Exam Questions

Let \(y = 3^x\). Then \(3^{-x} = \frac{1}{y}\).

The equation becomes \(\frac{y^2 + \frac{1}{y}}{y^2 - \frac{1}{y}} = 4\).

Multiply numerator and denominator by \(y\) to clear the fraction:

\(\frac{y^3 + 1}{y^3 - 1} = 4\).

Cross-multiply to get:

\(y^3 + 1 = 4(y^3 - 1)\).

Expand and simplify:

\(y^3 + 1 = 4y^3 - 4\).

Rearrange terms:

\(3y^3 = 5\).

So, \(y^3 = \frac{5}{3}\).

Since \(y = 3^x\), we have \((3^x)^3 = \frac{5}{3}\).

Thus, \(3^{3x} = \frac{5}{3}\).

Take the logarithm of both sides:

\(3x \log 3 = \log \left( \frac{5}{3} \right)\).

Solve for \(x\):

\(x = \frac{\log \left( \frac{5}{3} \right)}{3 \log 3}\).

Calculate \(x\) to 3 decimal places:

\(x \approx 0.155\).

First, express \(9^x\) in terms of base 3: \(9^x = (3^2)^x = (3^x)^2\).

Let \(u = 3^x\). Then the equation becomes \(u^2 = u + 12\).

Rearrange to form a quadratic equation: \(u^2 - u - 12 = 0\).

Factor the quadratic: \((u - 4)(u + 3) = 0\).

Thus, \(u = 4\) or \(u = -3\). Since \(u = 3^x\) and \(3^x > 0\), we have \(u = 4\).

Therefore, \(3^x = 4\).

Take the natural logarithm of both sides: \(x \ln 3 = \ln 4\).

Solve for \(x\): \(x = \frac{\ln 4}{\ln 3} \approx 1.26186\).

Round to two decimal places: \(x = 1.26\).

Let \(u = 5^x\). Then the equation becomes \(u^2 = u + 5\).

Rearrange to form a quadratic equation: \(u^2 - u - 5 = 0\).

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

Calculate the discriminant: \(b^2 - 4ac = (-1)^2 - 4 \times 1 \times (-5) = 1 + 20 = 21\).

Thus, \(u = \frac{1 \pm \sqrt{21}}{2}\).

Since \(u = 5^x\) must be positive, choose the positive root: \(u = \frac{1 + \sqrt{21}}{2}\).

Now, solve for \(x\): \(5^x = \frac{1 + \sqrt{21}}{2}\).

Take the logarithm of both sides: \(x \log 5 = \log \left(\frac{1 + \sqrt{21}}{2}\right)\).

Thus, \(x = \frac{\log \left(\frac{1 + \sqrt{21}}{2}\right)}{\log 5}\).

Calculate \(x\) to 3 decimal places: \(x \approx 0.638\).

Start with the equation:

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

Cross-multiply to eliminate the fraction:

\(3^x + 2 = 8(3^x - 2)\)

Expand the right side:

\(3^x + 2 = 8 \times 3^x - 16\)

Rearrange to isolate terms involving \(3^x\):

\(3^x + 2 = 8 \times 3^x - 16\)

\(2 + 16 = 8 \times 3^x - 3^x\)

\(18 = 7 \times 3^x\)

Divide both sides by 7:

\(3^x = \frac{18}{7}\)

Take the logarithm of both sides:

\(x \log 3 = \log \left( \frac{18}{7} \right)\)

Solve for \(x\):

\(x = \frac{\log \left( \frac{18}{7} \right)}{\log 3}\)

Calculate \(x\) to 3 decimal places:

\(x \approx 0.860\)

Substitute \(u = 3^x\) into the equation:

\(u + u^2 = u^3\).

Rearrange to form a quadratic equation:

\(u^3 - u^2 - u = 0\).

Factor out \(u\):

\(u(u^2 - u - 1) = 0\).

Since \(u = 3^x\) cannot be zero, solve \(u^2 - u - 1 = 0\).

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

\(u = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}\).

Select the positive root: \(u = \frac{1 + \sqrt{5}}{2}\).

Convert back to \(x\) using \(u = 3^x\):

\(3^x = \frac{1 + \sqrt{5}}{2}\).

Take the logarithm of both sides:

\(x \log 3 = \log \left( \frac{1 + \sqrt{5}}{2} \right)\).

Solve for \(x\):

\(x = \frac{\log \left( \frac{1 + \sqrt{5}}{2} \right)}{\log 3}\).

Calculate \(x\) to 3 significant figures:

\(x \approx 0.438\).

Start with the substitution \(u = 4^x\). The equation becomes:

\(u + 16 = 16u\)

Rearrange to solve for \(u\):

\(16u - u = 16\)

\(15u = 16\)

\(u = \frac{16}{15}\)

Since \(u = 4^x\), we have:

\(4^x = \frac{16}{15}\)

Take the logarithm of both sides:

\(x \log(4) = \log\left(\frac{16}{15}\right)\)

\(x = \frac{\log\left(\frac{16}{15}\right)}{\log(4)}\)

Calculate \(x\) to 3 significant figures:

\(x \approx 0.0466\)

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

Rewrite \(5^{x-1}\) as \(\frac{5^x}{5}\), giving \(\frac{5^x}{5} = 5^x - 5\).

Multiply through by 5 to clear the fraction: \(5^x = 5 \cdot (5^x - 5)\).

This simplifies to \(5^x = 5^{x+1} - 25\).

Rearrange to get \(5^{x+1} - 5^x = 25\).

Factor out \(5^x\): \(5^x(5 - 1) = 25\).

This simplifies to \(4 \cdot 5^x = 25\).

Divide by 4: \(5^x = \frac{25}{4}\).

Take the logarithm of both sides: \(x \ln 5 = \ln \left(\frac{25}{4}\right)\).

Solve for \(x\): \(x = \frac{\ln \left(\frac{25}{4}\right)}{\ln 5}\).

Calculate \(x\) to 3 significant figures: \(x \approx 1.14\).

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 \(1.83 < x < 1.95\).