Use natural logarithms to solve the equation; round to the nearest thousandth: \(3e^{2x} + 5 = 26\).

Step 1: Subtract 5 from both sides. \[3e^{2x} + 5 - 5 = 26 - 5\] \[3e^{2x} = 21\]

Step 2: Divide both sides by 3 to isolate \(e^{2x}\). \[\frac{3e^{2x}}{3} = \frac{21}{3}\] \[e^{2x} = 7\]

Step 3: Take the natural logarithm of both sides to "undo" the exponential. The natural logarithm of \(e\) to any power is simply that power, according to the property \(\ln(e^x) = x\). \[\ln(e^{2x}) = \ln(7)\] \[2x = \ln(7)\]

Step 4: Divide both sides by 2 to solve for \(x\). \[\frac{2x}{2} = \frac{\ln(7)}{2}\] \[x = \frac{\ln(7)}{2}\]

Step 5: Now use a calculator to evaluate the natural logarithm of 7 and divide it by 2, rounding to the nearest thousandth. \[x \approx \frac{\ln(7)}{2} \approx \frac{1.9459}{2} \approx 0.973\]

Thus, the solution to the equation \(3e^{2x} + 5 = 26\) is \(x \approx 0.973\), when rounded to the nearest thousandth.