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:

\(x = 4.30\).