Past Exam Questions

(a) Start with the equation \(\log_3(2x + 1) = 1 + 2\log_3(x - 1)\).

Using the law of logarithms, \(2\log_3(x - 1) = \log_3((x - 1)^2)\).

Thus, the equation becomes \(\log_3(2x + 1) = \log_3(3) + \log_3((x - 1)^2)\).

This simplifies to \(\log_3(2x + 1) = \log_3(3(x - 1)^2)\).

Equating the arguments gives \(2x + 1 = 3(x - 1)^2\).

Expanding \(3(x - 1)^2\) gives \(3(x^2 - 2x + 1) = 3x^2 - 6x + 3\).

Thus, \(2x + 1 = 3x^2 - 6x + 3\).

Rearranging gives \(3x^2 - 8x + 2 = 0\).

(b) Use the quadratic equation from part (a) to solve for \(y\):

\(3(2y)^2 - 8(2y) + 2 = 0\) simplifies to \(12y^2 - 16y + 2 = 0\).

Divide through by 2: \(6y^2 - 8y + 1 = 0\).

Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 6, b = -8, c = 1\).

\(y = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 6 \cdot 1}}{12}\).

\(y = \frac{8 \pm \sqrt{64 - 24}}{12}\).

\(y = \frac{8 \pm \sqrt{40}}{12}\).

\(y = \frac{8 \pm 2\sqrt{10}}{12}\).

\(y = \frac{4 \pm \sqrt{10}}{6}\).

Calculating gives \(y \approx 1.19\) (to 2 decimal places).

Start by using the property \(\log_{10} 10 = 1\) and \(\log_{10} 10^{-1} = -1\).

Rearrange the equation:

\(\log_{10}(2x + 1) = 2\log_{10}(x + 1) - 1\)

\(\log_{10}(2x + 1) + 1 = 2\log_{10}(x + 1)\)

\(\log_{10}(10(2x + 1)) = \log_{10}((x + 1)^2)\)

Equate the arguments:

\(10(2x + 1) = (x + 1)^2\)

Expand and simplify:

\(20x + 10 = x^2 + 2x + 1\)

\(x^2 - 18x - 9 = 0\)

Solve the quadratic equation using the quadratic formula:

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

where \(a = 1, b = -18, c = -9\).

\(x = \frac{18 \pm \sqrt{324 + 36}}{2}\)

\(x = \frac{18 \pm \sqrt{360}}{2}\)

\(x = \frac{18 \pm 18.9737}{2}\)

\(x = 18.487\) and \(x = -0.487\)

(i) Start with the equation \(\log_{10}(x-4) = 2 - \log_{10} x\).

Using the law of logarithms for subtraction, \(\log_{10}(x-4) + \log_{10} x = 2\).

This can be rewritten as \(\log_{10}((x-4)x) = 2\).

Since \(\log_{10} 100 = 2\), we have \((x-4)x = 100\).

Expanding gives \(x^2 - 4x = 100\).

Rearranging, we obtain the quadratic equation \(x^2 - 4x - 100 = 0\).

(ii) Solve the quadratic equation \(x^2 - 4x - 100 = 0\).

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -4\), and \(c = -100\).

Calculate the discriminant: \(b^2 - 4ac = (-4)^2 - 4 \times 1 \times (-100) = 16 + 400 = 416\).

Thus, \(x = \frac{4 \pm \sqrt{416}}{2}\).

Calculate \(\sqrt{416} \approx 20.396\).

So, \(x = \frac{4 \pm 20.396}{2}\).

This gives solutions \(x = 12.198\) and \(x = -8.198\).

Since \(x\) must be positive, the solution is \(x = 12.2\) (to 3 significant figures).

Start with the equation: \(2\log_2 x = 3 + \log_2(x + 1)\).

Apply the power rule of logarithms: \(\log_2 x^2 = 3 + \log_2(x + 1)\).

Use the property \(\log_a b - \log_a c = \log_a \left(\frac{b}{c}\right)\):

\(\log_2 \left(\frac{x^2}{x+1}\right) = 3\).

Convert the logarithmic equation to an exponential equation:

\(\frac{x^2}{x+1} = 2^3 = 8\).

Multiply both sides by \(x+1\):

\(x^2 = 8(x + 1)\).

Expand and rearrange to form a quadratic equation:

\(x^2 - 8x - 8 = 0\).

Solve the quadratic equation using the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -8\), \(c = -8\).

Calculate the discriminant: \(b^2 - 4ac = (-8)^2 - 4(1)(-8) = 64 + 32 = 96\).

\(x = \frac{8 \pm \sqrt{96}}{2}\).

\(x = \frac{8 \pm 4\sqrt{6}}{2}\).

\(x = 4 \pm 2\sqrt{6}\).

Calculate the positive root: \(x = 4 + 2\sqrt{6} \approx 8.90\).

Thus, the solution is \(x = 8.90\).

Start with the equation \(\log_{10}(x+9) = 2 + \log_{10} x\).

Use the logarithm property: \(\log_{10}(x+9) = \log_{10}(10^2) + \log_{10} x\).

This simplifies to \(\log_{10}(x+9) = \log_{10}(100x)\).

Since the logarithms are equal, set the arguments equal: \(x+9 = 100x\).

Rearrange to solve for \(x\):

\(x + 9 = 100x\)

\(9 = 99x\)

\(x = \frac{9}{99} = \frac{1}{11}\).

(i) Start with the equation \(\log_2(x+5) = 5 - \log_2 x\).

Using the law of logarithms, \(\log_2(x+5) + \log_2 x = 5\).

This can be rewritten as \(\log_2((x+5)x) = 5\).

Therefore, \((x+5)x = 2^5\).

So, \(x^2 + 5x = 32\).

Rearranging gives the quadratic equation \(x^2 + 5x - 32 = 0\).

(ii) Solve the quadratic equation \(x^2 + 5x - 32 = 0\).

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 5\), \(c = -32\).

Calculate the discriminant: \(b^2 - 4ac = 5^2 - 4 \times 1 \times (-32) = 25 + 128 = 153\).

Thus, \(x = \frac{-5 \pm \sqrt{153}}{2}\).

The positive solution is \(x = \frac{\sqrt{153} - 5}{2}\), which is approximately \(x = 3.68\).

(i) Start with the equation \(\log_{10}(x+5) = 2 - \log_{10} x\).

Use the property of logarithms: \(\log_{10} a - \log_{10} b = \log_{10} \left( \frac{a}{b} \right)\).

Rearrange the equation: \(\log_{10}(x+5) + \log_{10} x = 2\).

Combine the logarithms: \(\log_{10}((x+5)x) = 2\).

Convert the logarithmic equation to an exponential equation: \((x+5)x = 10^2\).

Simplify: \(x^2 + 5x = 100\).

Thus, the quadratic equation is \(x^2 + 5x - 100 = 0\).

(ii) Solve the quadratic equation \(x^2 + 5x - 100 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 5\), and \(c = -100\).

Calculate the discriminant: \(b^2 - 4ac = 5^2 - 4 \times 1 \times (-100) = 25 + 400 = 425\).

Find the roots: \(x = \frac{-5 \pm \sqrt{425}}{2}\).

Since \(x\) must be positive, choose the positive root: \(x = \frac{\sqrt{425} - 5}{2}\).

Approximate the value: \(x \approx 7.81\).

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

Rewrite \(4^{x-2}\) as \(\frac{4^x}{4^2}\), so the equation becomes:

\(\frac{4^x}{16} = 4^x - 16\).

Multiply through by 16 to clear the fraction:

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

Expand the right side:

\(4^x = 16 \cdot 4^x - 256\).

Rearrange to isolate terms involving \(4^x\):

\(4^x - 16 \cdot 4^x = -256\).

Factor out \(4^x\):

\(4^x(1 - 16) = -256\).

Simplify:

\(-15 \cdot 4^x = -256\).

Divide both sides by -15:

\(4^x = \frac{256}{15}\).

Take the logarithm of both sides to solve for \(x\):

\(x \log(4) = \log\left(\frac{256}{15}\right)\).

Solve for \(x\):

\(x = \frac{\log\left(\frac{256}{15}\right)}{\log(4)}\).

Calculate the value:

\(x \approx 2.047\).

Start with the equation:

\(\frac{2^x + 1}{2^x - 1} = 5\)

Multiply both sides by \(2^x - 1\):

\(2^x + 1 = 5(2^x - 1)\)

Expand the right side:

\(2^x + 1 = 5 \cdot 2^x - 5\)

Rearrange the terms:

\(2^x + 1 = 5 \cdot 2^x - 5\)

\(1 + 5 = 5 \cdot 2^x - 2^x\)

\(6 = 4 \cdot 2^x\)

Divide both sides by 4:

\(2^x = \frac{6}{4} = 1.5\)

Take the logarithm base 2 of both sides:

\(x = \log_2(1.5)\)

Calculate \(x\) using a calculator:

\(x \approx 0.585\)

Thus, the solution is \(x = 0.585\) to 3 significant figures.

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

Rewrite \(3^{x+2}\) as \(3^x imes 3^2\), giving \(3^x imes 9 = 3^x + 9\).

Factor out \(3^x\) from the left side: \(3^x (9 - 1) = 9\).

This simplifies to \(3^x imes 8 = 9\).

Divide both sides by 8: \(3^x = \frac{9}{8}\).

Take the logarithm of both sides: \(x \ln 3 = \ln \left(\frac{9}{8}\right)\).

Solve for \(x\): \(x = \frac{\ln \left(\frac{9}{8}\right)}{\ln 3}\).

Calculate \(x\) to 3 significant figures: \(x \approx 0.107\).

Start by substituting \(u = 3^x\). Then \(3^{-x} = \frac{1}{3^x} = \frac{1}{u}\).

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

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

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

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

Calculate the discriminant: \(b^2 - 4ac = (-2)^2 - 4(1)(-1) = 4 + 4 = 8\).

Thus, \(u = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}\).

Since \(u = 3^x\) must be positive, choose \(u = 1 + \sqrt{2}\).

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

Take the logarithm of both sides: \(x \log 3 = \log(1 + \sqrt{2})\).

Thus, \(x = \frac{\log(1 + \sqrt{2})}{\log 3}\).

Calculate \(x \approx 0.802\) to 3 significant figures.

(i) Given \(y = 2^x\), we have \(2^{-x} = \frac{1}{y}\). Substitute into the equation:

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

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

\(y^2 - 1 = y\).

Rearrange to form a quadratic equation:

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

(ii) Solve the quadratic equation \(y^2 - y - 1 = 0\) using the quadratic formula:

\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = -1, c = -1\).

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

Since \(y = 2^x\) and \(y > 0\), choose the positive root:

\(y = \frac{1 + \sqrt{5}}{2}\).

Now solve \(2^x = \frac{1 + \sqrt{5}}{2}\) by taking logarithms:

\(x = \log_2 \left( \frac{1 + \sqrt{5}}{2} \right)\).

Calculate \(x \approx 0.694\).

Let \(u = 4^x\). Then \(4^{-x} = \frac{1}{u}\).

The equation becomes \(u = 3 + \frac{1}{u}\).

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

Rearrange to form a quadratic equation: \(u^2 - 3u - 1 = 0\).

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

\(u = \frac{3 \pm \sqrt{9 + 4}}{2} = \frac{3 \pm \sqrt{13}}{2}\).

Since \(u = 4^x\) must be positive, take the positive root: \(u = \frac{3 + \sqrt{13}}{2}\).

Calculate \(u \approx 3.30277563773\).

Since \(u = 4^x\), solve for \(x\): \(x = \log_4(u)\).

\(x = \frac{\log(u)}{\log(4)}\).

\(x \approx \frac{\log(3.30277563773)}{\log(4)} \approx 0.862\).

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: \(x \approx 1.14\).