Tasks 02-05 - Exponential, Logarithmic & Complex Word Problems
Section 02: Equations & Problem-Solving Strategies
Instructions
Complete these problems to master advanced exponential and logarithmic equations, as well as complex multi-step word problems. These problems integrate all equation-solving techniques from Section 02.
Problem 1: Exponential Equations (x)
Solve each exponential equation:
\(2^{x+3} = 128\)
\(5^x \cdot 25^{x-1} = 125\)
\(3^{2x} - 12 \cdot 3^x + 27 = 0\)
\(4^x - 2^{x+1} - 8 = 0\)
Problem 2: Logarithmic Equations (x)
Solve each logarithmic equation, stating domain restrictions:
\(\log_3(x + 4) = 2\)
\(\log(x) + \log(x + 3) = 1\)
\(\log_2(x) - \log_8(x) = 1\)
\(\log_x(16) = 4\)
Problem 3: Mixed Exponential-Logarithmic (xx)
Solve these equations that involve both exponential and logarithmic expressions:
\(3^{\log_3(x)} + x = 10\)
\(\log_2(2^x + 1) = x - 1\)
Note, there might be multiple solutions or no solution at all!
Problem 4: Growth and Decay Applications (xx)
A radioactive substance decays according to \(A(t) = A_0 e^{-kt}\) where \(t\) is in years.
- If 70% remains after 5 years, find the decay constant \(k\)
- Find the half-life of the substance
- How long until only 10% remains?
- If we start with 100 grams, when will exactly 25 grams remain?
Problem 5: Investment Comparison (xx)
An investor has €30,000 to invest. She’s considering three options:
- Option A: 6% annual interest, compounded yearly
- Option B: 5.8% annual interest, compounded monthly
- Option C: 5.7% annual interest, compounded continuously
- Write the formula for each option after \(t\) years
- Which option yields the most after 10 years?
- If she wants €50,000, how long would each option take?
Problem 6: Population Dynamics (xxx)
Two bacterial cultures are growing in a lab. Culture A starts with 500 bacteria and doubles every 3 hours. Culture B starts with 800 bacteria and grows according to \(B(t) = 800e^{0.15t}\) where \(t\) is in hours.
- Write the growth equation for Culture A
- When will the populations be equal?