We are given the equation:

$\frac{30}{1 + 7e^{-0.7t}} = 10$

We need to solve for $t$ using logarithms. Here's the step-by-step solution:

Step 1: Multiply both sides by $1 + 7e^{-0.7t}$ to eliminate the denominator.

$30 = 10(1 + 7e^{-0.7t})$

Step 2: Expand the right-hand side.

$30 = 10 + 70e^{-0.7t}$

Step 3: Subtract 10 from both sides.

$20 = 70e^{-0.7t}$

Step 4: Divide both sides by 70 to isolate the exponential term.

$\frac{20}{70} = e^{-0.7t}$ $\frac{2}{7} = e^{-0.7t}$

Step 5: Take the natural logarithm (ln) of both sides to eliminate the exponential.

$\ln\left(\frac{2}{7}\right) = -0.7t$

Step 6: Solve for $t$.

$t = \frac{\ln\left(\frac{2}{7}\right)}{-0.7}$

Step 7: Compute the natural logarithm and divide.

$t = \frac{\ln(0.2857)}{-0.7} \approx \frac{-1.25276}{-0.7} \approx 1.79$

Thus, the solution is:

$t \approx 1.79$

Would you like further details or explanations?

Here are 5 related questions:

1. How do logarithmic properties simplify solving exponential equations?
2. What is the importance of natural logarithms in solving exponential functions?
3. Can the solution change if we use a different base for the logarithm?
4. How do we interpret the value of $t$ in real-world exponential models?
5. What happens if we have a different constant in the exponent?

Tip: Always check your work by substituting the solution back into the original equation to verify its correctness.