25+150e^-(ln3/2)t/20 = 25 + 250e^-ln(5/3)t/20

We need to solve the equation:

\[
25 + 150e^{-\frac{\ln{3}}{2} \cdot \frac{t}{20}} = 25 + 250e^{-\ln{\frac{5}{3}} \cdot \frac{t}{20}}
\]

### Step-by-Step Solution:

1. Subtract 25 from both sides:
   \[
   150e^{-\frac{\ln{3}}{2} \cdot \frac{t}{20}} = 250e^{-\ln{\frac{5}{3}} \cdot \frac{t}{20}}
   \]

2. Divide both sides by 150:
   \[
   e^{-\frac{\ln{3}}{2} \cdot \frac{t}{20}} = \frac{250}{150} e^{-\ln{\frac{5}{3}} \cdot \frac{t}{20}}
   \]
   Simplifying the fraction:
   \[
   e^{-\frac{\ln{3}}{2} \cdot \frac{t}{20}} = \frac{5}{3} e^{-\ln{\frac{5}{3}} \cdot \frac{t}{20}}
   \]

3. Now, take the natural logarithm of both sides:
   \[
   -\frac{\ln{3}}{2} \cdot \frac{t}{20} = \ln{\frac{5}{3}} - \ln{\frac{5}{3}} \cdot \frac{t}{20}
   \]

4. Combine like terms:
   \[
   -\frac{\ln{3}}{2} \cdot \frac{t}{20} + \ln{\frac{5}{3}} \cdot \frac{t}{20} = \ln{\frac{5}{3}}
   \]

5. Factor out \(\frac{t}{20}\):
   \[
   \frac{t}{20} \left( \ln{\frac{5}{3}} - \frac{\ln{3}}{2} \right) = \ln{\frac{5}{3}}
   \]

6. Solve for \( t \):
   \[
   t = \frac{20 \ln{\frac{5}{3}}}{\ln{\frac{5}{3}} - \frac{\ln{3}}{2}}
   \]

Now, you can simplify this further by calculating the values of the logarithms if needed.

Would you like further details on any step or help with simplifying the final expression?

Here are some related questions:
1. How do you simplify expressions with logarithms and exponents?
2. What properties of logarithms are used to solve exponential equations?
3. Can you always take the natural logarithm on both sides of an equation?
4. What are some real-world applications of equations involving exponential decay?
5. How can you graph equations involving exponential functions?

**Tip:** Always check your final expression for any simplifications involving common logarithmic identities!

25 + 150e^-(ln3/2)t/20 = 25 + 250e^-ln(5/3)t/20

Learn how to solve the equation 25 + 150e^-(ln3/2)t/20 = 25 + 250e^-ln(5/3)t/20 using logarithmic identities and exponential decay concepts. This step-by-step guide includes key formulas and theorems, ideal for students in Grades 10-12.

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