Past Exam Questions

Start with the equation \(2(3^{2x-1}) = 4^{x+1}\).

Take the logarithm of both sides:

\(\ln(2) + (2x-1)\ln(3) = (x+1)\ln(4)\).

Rearrange to form a linear equation:

\((2x-1)\ln(3) - x\ln(4) = \ln(4) - \ln(2)\).

Simplify and solve for \(x\):

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

Calculate the value of \(x\) to 2 decimal places:

\(x \approx 2.21\).

Start with the equation \(3(2^{1-x}) = 7^x\).

Take the natural logarithm of both sides: \(\ln(3(2^{1-x})) = \ln(7^x)\).

Using the logarithm properties, this becomes \(\ln 3 + (1-x)\ln 2 = x\ln 7\).

Rearrange to form a linear equation: \(\ln 3 + \ln 2 - x\ln 2 = x\ln 7\).

Combine terms: \(\ln 3 + \ln 2 = x\ln 7 + x\ln 2\).

Factor out \(x\): \(\ln 3 + \ln 2 = x(\ln 7 + \ln 2)\).

Solve for \(x\): \(x = \frac{\ln 3 + \ln 2}{\ln 7 + \ln 2}\).

Simplify: \(x = \frac{\ln 6}{\ln 14}\).

Start with the given equation:

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

Apply the logarithm property \(a \ln b = \ln(b^a)\):

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

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

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

Since the logarithms are equal, set the arguments equal:

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

Multiply through by 3 to clear the fraction:

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

\(3x^3 - 9 = x^3\)

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

\(3x^3 - x^3 = 9\)

\(2x^3 = 9\)

Divide by 2:

\(x^3 = \frac{9}{2}\)

Take the cube root:

\(x = \sqrt[3]{\frac{9}{2}}\)

Calculate \(x\) to 3 significant figures:

\(x \approx 1.65\)

Let \(y = e^{2x}\). Then \(e^{-2x} = \frac{1}{y}\).

The equation becomes:

\(3y - \frac{4}{y} = 5\).

Multiply through by \(y\) to clear the fraction:

\(3y^2 - 5y - 4 = 0\).

This is a quadratic equation in \(y\). Solve using the quadratic formula:

\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -5\), \(c = -4\).

\(y = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3}\)

\(y = \frac{5 \pm \sqrt{25 + 48}}{6}\)

\(y = \frac{5 \pm \sqrt{73}}{6}\)

Since \(y = e^{2x}\), we have:

\(e^{2x} = \frac{5 + \sqrt{73}}{6}\)

Take the natural logarithm of both sides:

\(2x = \ln\left(\frac{5 + \sqrt{73}}{6}\right)\)

\(x = \frac{1}{2} \ln\left(\frac{5 + \sqrt{73}}{6}\right)\)

Calculating this gives:

\(x \approx 0.407\)

First, let \(u = e^x\). Then the equation becomes \(\frac{2u + \frac{1}{u}}{2 + u} = 3\).

Multiply through by \(u(2 + u)\) to clear the fraction: \(2u^2 + 1 = 3u(2 + u)\).

Expand and simplify: \(2u^2 + 1 = 6u + 3u^2\).

Rearrange to form a quadratic equation: \(u^2 + 6u - 1 = 0\).

Use the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1, b = 6, c = -1\).

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

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

Simplify: \(u = -3 \pm \sqrt{10}\).

Since \(u = e^x > 0\), we reject \(u = -3 - \sqrt{10}\) as it is negative.

Thus, \(u = -3 + \sqrt{10}\).

Find \(x\) by taking the natural logarithm: \(x = \ln(-3 + \sqrt{10})\).

Calculate \(x\) to 3 decimal places: \(x = -1.818\).

To show there is only one real root, note that the quadratic equation \(u^2 + 6u - 1 = 0\) has only one positive solution for \(u\), which corresponds to one real solution for \(x\).

(a) Start with the equation \(\ln(1 + e^{-x}) + 2x = 0\).

Remove the logarithm: \(1 + e^{-x} = e^{-2x}\).

Substitute \(u = e^x\), then \(e^{-x} = \frac{1}{u}\) and \(e^{-2x} = \frac{1}{u^2}\).

The equation becomes \(1 + \frac{1}{u} = \frac{1}{u^2}\).

Multiply through by \(u^2\) to clear fractions: \(u^2 + u - 1 = 0\).

(b) Solve the quadratic equation \(u^2 + u - 1 = 0\).

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

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

Find the roots: \(u = \frac{-1 \pm \sqrt{5}}{2}\).

Choose the positive root: \(u = \frac{-1 + \sqrt{5}}{2} \approx 0.618\).

Since \(u = e^x\), solve for \(x\): \(x = \ln(0.618)\).

Calculate \(x \approx -0.481\).

Let \(u = e^x\). Then \(e^{-x} = \frac{1}{u}\).

The equation becomes:

\(\frac{u + \frac{1}{u}}{u + 1} = 4\).

Multiply through by \(u(u + 1)\) to clear the fraction:

\(u^2 + 1 = 4u(u + 1)\).

Expand and simplify:

\(u^2 + 1 = 4u^2 + 4u\).

Rearrange to form a quadratic equation:

\(3u^2 + 4u - 1 = 0\).

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

\(u = \frac{-4 \pm \sqrt{4^2 - 4 \times 3 \times (-1)}}{2 \times 3}\).

\(u = \frac{-4 \pm \sqrt{16 + 12}}{6}\).

\(u = \frac{-4 \pm \sqrt{28}}{6}\).

\(u = \frac{-4 \pm 2\sqrt{7}}{6}\).

\(u = \frac{-2 \pm \sqrt{7}}{3}\).

Choose the positive root: \(u = \frac{\sqrt{7} - 2}{3}\).

Since \(u = e^x\), solve for \(x\):

\(x = \ln\left(\frac{\sqrt{7} - 2}{3}\right)\).

Calculate \(x\) to 3 decimal places:

\(x \approx -1.536\).

Start with the equation:

\(\frac{2e^x + e^{-x}}{e^x - e^{-x}} = 4\)

Multiply both sides by \(e^x - e^{-x}\):

\(2e^x + e^{-x} = 4(e^x - e^{-x})\)

Expand the right side:

\(2e^x + e^{-x} = 4e^x - 4e^{-x}\)

Rearrange terms:

\(2e^x + e^{-x} + 4e^{-x} = 4e^x\)

\(2e^x + 5e^{-x} = 4e^x\)

Rearrange to isolate terms with \(e^x\):

\(2e^x - 4e^x = -5e^{-x}\)

\(-2e^x = -5e^{-x}\)

Divide both sides by \(-2\):

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

Multiply both sides by \(e^x\):

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

Take the natural logarithm of both sides:

\(2x = \ln\left(\frac{5}{2}\right)\)

Solve for \(x\):

\(x = \frac{1}{2} \ln\left(\frac{5}{2}\right)\)

Calculate \(x\) to 2 decimal places:

\(x \approx 0.46\)

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

\(4e^{-x} = 4 \cdot \frac{1}{e^x} = \frac{4}{u}\).

The equation becomes \(\frac{4}{u} = 3u + 4\).

Multiply through by \(u\) to clear the fraction:

\(4 = 3u^2 + 4u\).

Rearrange to form a quadratic equation:

\(3u^2 + 4u - 4 = 0\).

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

\(u = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3}\).

\(u = \frac{-4 \pm \sqrt{16 + 48}}{6}\).

\(u = \frac{-4 \pm \sqrt{64}}{6}\).

\(u = \frac{-4 \pm 8}{6}\).

Solutions for \(u\) are \(u = \frac{4}{6} = \frac{2}{3}\) or \(u = \frac{-12}{6} = -2\).

Since \(u = e^x\) must be positive, we take \(u = \frac{2}{3}\).

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

Take the natural logarithm of both sides:

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

Calculate \(x\) to 3 significant figures:

\(x \approx -0.405\).

Start with the equation \(e^x = 1 + 6e^{-x}\).

Multiply both sides by \(e^x\) to eliminate the negative exponent: \(e^{2x} = e^x + 6\).

Rearrange to form a quadratic equation: \(e^{2x} - e^x - 6 = 0\).

Let \(u = e^x\), then the equation becomes \(u^2 - u - 6 = 0\).

Factor the quadratic: \((u - 3)(u + 2) = 0\).

Thus, \(u = 3\) or \(u = -2\). Since \(u = e^x\) and \(e^x > 0\), \(u = -2\) is not valid.

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

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

Calculate \(x\) to 3 significant figures: \(x = 1.10\).

Start by rewriting the equation \(e^x + e^{2x} = e^{3x}\) as \(e^x + e^{2x} - e^{3x} = 0\).

Let \(u = e^x\), then the equation becomes \(u + u^2 - u^3 = 0\).

Factor the equation: \(u(u^2 + u - 1) = 0\).

Since \(u = e^x > 0\), we discard \(u = 0\) and solve \(u^2 + u - 1 = 0\).

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

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

Choose the positive root: \(u = \frac{1 + \sqrt{5}}{2} \approx 1.618\).

Since \(u = e^x\), solve for \(x\): \(x = \ln\left(\frac{1 + \sqrt{5}}{2}\right)\).

Calculate \(x \approx 0.481\) to 3 significant figures.

Start with the given equation:

\(x = \ln(2y - 3) - \ln(y + 4)\)

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

\(x = \ln \left( \frac{2y - 3}{y + 4} \right)\)

Exponentiate both sides to remove the logarithm:

\(e^x = \frac{2y - 3}{y + 4}\)

Cross-multiply to solve for \(y\):

\(e^x (y + 4) = 2y - 3\)

Expand and rearrange terms:

\(e^x y + 4e^x = 2y - 3\)

\(2y - e^x y = 3 + 4e^x\)

Factor out \(y\):

\(y(2 - e^x) = 3 + 4e^x\)

Solve for \(y\):

\(y = \frac{3 + 4e^x}{2 - e^x}\)

Start with the equation \(\ln(1 + e^{2y}) = x\).

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

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

Take the natural logarithm of both sides: \(2y = \ln(e^x - 1)\).

Divide by 2 to solve for \(y\): \(y = \frac{1}{2} \ln(e^x - 1)\).

Start with the given equation:

\(z = \ln(y+2) - \ln(y+1)\)

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

Thus, \(z = \ln\left(\frac{y+2}{y+1}\right)\).

Exponentiate both sides to remove the logarithm:

\(e^z = \frac{y+2}{y+1}\)

Cross-multiply to solve for \(y\):

\(e^z(y+1) = y+2\)

\(e^z y + e^z = y + 2\)

Rearrange terms to isolate \(y\):

\(e^z y - y = 2 - e^z\)

\(y(e^z - 1) = 2 - e^z\)

\(y = \frac{2 - e^z}{e^z - 1}\)

Start with the equation:

\(2 \ln(x + 4) - \ln x = \ln(x + a)\)

Apply the logarithm property \(a \ln b = \ln(b^a)\):

\(\ln((x + 4)^2) - \ln x = \ln(x + a)\)

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

\(\ln\left(\frac{(x + 4)^2}{x}\right) = \ln(x + a)\)

Since the logarithms are equal, set the arguments equal:

\(\frac{(x + 4)^2}{x} = x + a\)

Clear the fraction by multiplying through by \(x\):

\((x + 4)^2 = x(x + a)\)

Expand both sides:

\(x^2 + 8x + 16 = x^2 + ax\)

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

\(8x + 16 = ax\)

Rearrange to solve for \(x\):

\(ax - 8x = 16\)

\(x(a - 8) = 16\)

Divide by \(a - 8\):

\(x = \frac{16}{a - 8}\)

Start with the given equation:

\(\ln(y + 1) - \ln y = 1 + 3 \ln x\)

Use the logarithm quotient rule: \(\ln \left( \frac{y+1}{y} \right) = 1 + 3 \ln x\)

Exponentiate both sides to remove the logarithm:

\(\frac{y+1}{y} = e^{1 + 3 \ln x}\)

Use the property \(e^{a+b} = e^a \cdot e^b\):

\(\frac{y+1}{y} = e \cdot (e^{\ln x})^3\)

Since \(e^{\ln x} = x\), this becomes:

\(\frac{y+1}{y} = e \cdot x^3\)

Rearrange to solve for \(y\):

\(y + 1 = e x^3 y\)

\(1 = e x^3 y - y\)

\(1 = y(e x^3 - 1)\)

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

Thus, \(y = (e^{x^3} - 1)^{-1}\).

Start with the equation \(x = 4(3^{-y})\).

Take the natural logarithm of both sides: \(\ln x = \ln(4 \cdot 3^{-y})\).

Apply the logarithm of a product: \(\ln x = \ln 4 + \ln(3^{-y})\).

Use the logarithm of a power: \(\ln(3^{-y}) = -y \ln 3\).

Substitute back: \(\ln x = \ln 4 - y \ln 3\).

Rearrange to solve for \(y\): \(y = \frac{\ln 4 - \ln x}{\ln 3}\).

Answer: \(a\approx4.06\), \(b\approx9.97\).

Given

\(y=ab^x,\)

take natural logarithms:

\(\ln y=\ln a+x\ln b.\)

The graph of \(\ln y\) against \(x\) is therefore a straight line with gradient \(\ln b\) and intercept \(\ln a\).

The two given points are \((1,3.7)\) and \((2.2,6.46)\), so

\(3.7=\ln a+\ln b\)

and

\(6.46=\ln a+2.2\ln b.\)

Subtracting the first equation from the second gives

\(2.76=1.2\ln b.\)

Thus

\(\ln b=2.3\)

and

\(b=e^{2.3}\approx9.97.\)

Substitute \(\ln b=2.3\) into \(3.7=\ln a+\ln b\):

\(\ln a=3.7-2.3=1.4.\)

Therefore

\(a=e^{1.4}\approx4.06.\)

(i) Substitute \(x = \ln 3\) and \(y = \ln 2\) into the equation:

\(6e^{2\ln 3} + ke^{\ln 2} + e^{2\ln 2} = c\).

Simplify using \(e^{2\ln 3} = 9\), \(e^{\ln 2} = 2\), and \(e^{2\ln 2} = 4\):

\(6 \times 9 + k \times 2 + 4 = c\).

\(54 + 2k + 4 = c\).

\(58 + 2k = c\).

(ii) Differentiate the left-hand side with respect to \(x\):

\(\frac{d}{dx}(6e^{2x} + ke^y + e^{2y}) = 0\).

\(12e^{2x} + ke^y \frac{dy}{dx} + 2e^{2y} \frac{dy}{dx} = 0\).

Substitute \(x = \ln 3\), \(y = \ln 2\), and \(\frac{dy}{dx} = -6\):

\(12 \times 9 + 2k(-6) + 2 \times 4(-6) = 0\).

\(108 - 12k - 48 = 0\).

\(60 = 12k\).

\(k = 5\).

Substitute \(k = 5\) into \(58 + 2k = c\):

\(58 + 2 \times 5 = c\).

\(c = 68\).

To solve part (i), start with the given equation:

\(y^3 = Ae^{2x}\)

Take the natural logarithm of both sides:

\(\ln(y^3) = \ln(Ae^{2x})\)

Using logarithm properties, this becomes:

\(3\ln(y) = \ln(A) + 2x\)

This is in the form \(y = mx + c\), where the gradient \(m\) is \(\frac{2}{3}\).

For part (ii), substitute \(x = 0\) and \(\ln(y) = 0.5\) into the equation:

\(3(0.5) = \ln(A) + 2(0)\)

\(1.5 = \ln(A)\)

Exponentiate both sides to solve for \(A\):

\(A = e^{1.5}\)

Calculate \(A\) to two decimal places:

\(A \approx 4.48\)