"e" (AQA A-Level Mathematics): Revision Notes

• If $y = e^{x}$, then $\frac{dy}{dx} = e^{x}$.

Example 1: Solve $e^{2x + 3} = 5$

1. Take the natural logarithm of both sides: $\ln(e^{2x + 3}) = \ln(5)$

2. Simplify: $2x + 3 = \ln(5)$

3. Solve for $x$: $2x = \ln(5) - 3$

$x = \frac{1}{2}(\ln(5) - 3)$

• The gradient function of $y = e^{kx}$ is given by: $\frac{dy}{dx} = ke^{kx}$

Example 2: Solve $e^{\ln(x^2)} = 25$

1. Simplify using the property $e^{\ln(x^2)} = x^2$: $x^2 = 25$

2. Solve for $x$: $x = \pm 5$

3. Since $\ln(-5)$ is undefined, the valid solution is: $x = 5$

• Note: It is important to consider whether answers are valid. When discarding an answer, state why.

Example: Solve $\ln(x - 4) = 7$

1. Start with the given equation: $\ln(x - 4) = 7$

2. Exponentiate both sides to eliminate the natural logarithm: $e^{\ln(x - 4)} = e^7$

3. Simplify: $x - 4 = e^7$

4. Solve for $x$: $x = e^7 + 4$

Example 3: Find the Time Taken for the Mass to Decrease to 25% of its Value When $t = 0$

1. At $t = 0$: $M = 40e^{-0.132(0)} = 40e^0 = 40$

2. Find 25% of the initial mass: $0.25 \times 40 = 10$

3. Set up the equation: $10 = 40e^{-0.132t}$

4. Divide both sides by 40: $\frac{10}{40} = e^{-0.132t}$

5. Take the natural logarithm of both sides: $\ln(0.25) = -0.132t$

6. Solve for $t$: $t = \frac{\ln(0.25)}{-0.132} \approx :highlight[10.5 years] (3\ significant\ figures)$

• $e$ is a fundamental constant in mathematics, used especially in exponential and logarithmic functions.
• The natural logarithm $\ln(x)$ is the inverse of the exponential function $e^{x}$.
• In solving exponential equations, it's crucial to consider the validity of solutions, particularly with logarithms.