Lecture Notes: Exponentials and Logarithms
Exponential Equations
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Key Principle: If two exponential expressions with the same base are equal, then their exponents are equal (i.e., if a^x = a^y, then x = y).
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Solving Steps:
- Isolate the exponential expression.
- Take the logarithm of both sides using the law of logarithms.
- Use the power rule to bring the exponent out front.
- Solve for the variable.
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Example:
- Given equation: 3^(x+2) = 7
- Steps:
- Take log of both sides.
- Bring (x + 2) out front using log rules.
- Solve: x = (log 7 / log 3) - 2.
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Checking solutions:
- Ensure precision by checking work, taking into account possible round-off errors.
Another Method: Substitution
- Example:
- Start with equation e^(2x) - e^x - 6 = 0.
- Substitute: u = e^x, then solve the quadratic u^2 - u - 6 = 0.
- Solutions: u = 3 or u = -2. Only u = 3 is valid (since exponential functions can't be negative).
- Solve: e^x = 3, so x = ln(3).
Logarithmic Equations
Applications of Logarithms
- Refraction of Light:
- Light refraction changes its direction when passing through different mediums.
- Beer-Lambert Law: Used to describe the intensity of light through a material.
- Equation: I = Iā * e^(-k * x), where k is a constant for the material.
- Example: Calculate light intensity at 20 feet with given parameters.
Compound Interest and Logarithms
- Example Problem: Finding time to double an investment at a 5% interest rate.
- Equations:
- Semi-Annual Compounding: A = P(1 + r/n)^(nt)
- Continuous Compounding: A = Pe^(rt)
- Solution:
- Semi-Annually: Takes ~14 years to double.
- Continuously: Takes ~13.86 years.
- Conclusion: Compounding more frequently has a small effect, but interest rate plays a more critical role in investment growth.
These notes summarize the concepts and examples discussed in the lecture on exponential and logarithmic equations, their solutions, and real-world applications through light refraction and compound interest.