Solve for t . \( e^{-0.49 \mathrm{t}}=0.70 \) Select the correct choice below and, if necessary, fill in the answer box to compl A. The solution is \( \mathrm{t}=\square \). Type an integer or a decimal. Do not round until the final answer. Then rou B. The solution is not a real number.

To solve for \( t \), we start by taking the natural logarithm of both sides: \[ \ln(e^{-0.49t}) = \ln(0.70) \] Using the property of logarithms that states \( \ln(e^x) = x \), we simplify the left side: \[ -0.49t = \ln(0.70) \] Now we can isolate \( t \) by dividing both sides by \( -0.49 \): \[ t = \frac{\ln(0.70)}{-0.49} \] Calculating \( \ln(0.70) \) gives us approximately \(-0.3567\): \[ t \approx \frac{-0.3567}{-0.49} \approx 0.727 \] So, the solution is: A. The solution is \( \mathrm{t} \approx 0.727 \). Type an integer or a decimal. Do not round until the final answer. Then round to the desired precision. Now you know how to tackle equations involving exponents and logarithms! Just remember that applying natural logarithms allows you to bring down the exponent and makes isolating the variable much easier. It’s like doing a magic trick with numbers! Need some real-world application? You encounter exponential decay in activities like radioactive decay or even in compound interest calculations. Understanding how to manipulate these equations can help you make predictions and analyze growth patterns in many fields, from finance to biology!