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 \(x = 3.62\).

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

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

Rearrange to solve for \(e^{-x}\): \(e^{-x} = e^2 - 2\).

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

Multiply by -1 to solve for \(x\): \(x = -\ln(e^2 - 2)\).

Calculate the value: \(x \approx -1.68\).

Start with the equation \(\ln(x+2) = 2 + \ln x\).

Use the properties of logarithms to combine the logarithms: \(\ln(x+2) - \ln x = 2\).

This simplifies to \(\ln\left(\frac{x+2}{x}\right) = 2\).

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

Multiply both sides by \(x\) to get \(x+2 = e^2 x\).

Rearrange to solve for \(x\): \(x - e^2 x = -2\).

Factor out \(x\): \(x(1 - e^2) = -2\).

Solve for \(x\): \(x = \frac{-2}{1 - e^2}\).

Calculate \(x\) to three decimal places: \(x \approx 0.313\).

Start with the equation \(\ln(1 + x) = 1 + \ln x\).

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

Rearrange the equation: \(\ln(1 + x) - \ln x = 1\).

This becomes \(\ln \left( \frac{1 + x}{x} \right) = 1\).

Exponentiate both sides to remove the logarithm: \(\frac{1 + x}{x} = e\).

Rearrange to solve for \(x\): \(1 + x = ex\).

\(1 = ex - x\).

\(1 = x(e - 1)\).

\(x = \frac{1}{e - 1}\).

Calculate \(x\) using \(e \approx 2.718\):

\(x \approx \frac{1}{2.718 - 1} \approx 0.582\).

Round to 2 significant figures: \(x = 0.58\).

Start with the equation:

\(\ln(2x - 1) = 2 \ln(x + 1) - \ln x\)

Apply the logarithm power rule: \(2 \ln(x + 1) = \ln((x + 1)^2)\)

So the equation becomes:

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

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

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

Since the logarithms are equal, the arguments must be equal:

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

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

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

Expand both sides:

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

Rearrange to form a quadratic equation:

\(2x^2 - x - x^2 - 2x - 1 = 0\)

\(x^2 - 3x - 1 = 0\)

Solve the quadratic equation using the quadratic formula:

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

Here, \(a = 1\), \(b = -3\), \(c = -1\)

\(x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1}\)

\(x = \frac{3 \pm \sqrt{9 + 4}}{2}\)

\(x = \frac{3 \pm \sqrt{13}}{2}\)

Calculate the positive root:

\(x = \frac{3 + \sqrt{13}}{2} \approx 3.303\)

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

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

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

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

Expand and rearrange the equation:

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

Combine like terms:

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

Factor out \(x\):

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

Solve for \(x\):

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

Combine the logarithms:

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

\(x = \frac{\ln 30}{\ln 24}\)

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

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

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

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

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

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

Simplify the expression: \(x = \frac{\ln 10}{\ln 24}\).

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

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

Rearrange the equation: \(\ln(e^{2x} + 3) - \ln 3 = 2x\).

This becomes \(\ln\left(\frac{e^{2x} + 3}{3}\right) = 2x\).

Exponentiate both sides to remove the logarithm: \(\frac{e^{2x} + 3}{3} = e^{2x}\).

Multiply through by 3: \(e^{2x} + 3 = 3e^{2x}\).

Rearrange to form a quadratic in terms of \(e^{2x}\): \(3 = 2e^{2x}\).

Divide by 2: \(e^{2x} = \frac{3}{2}\).

Take the natural logarithm of both sides: \(2x = \ln\left(\frac{3}{2}\right)\).

Solve for \(x\): \(x = \frac{1}{2} \ln\left(\frac{3}{2}\right)\).

Calculate \(x\) to 3 decimal places: \(x \approx 0.203\).

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 \(y\): \(y = \frac{\ln 4 - \ln x}{\ln 3}\).