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 \(y\):

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

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

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