Exponential Functions

written as $a b^{x}$ where
- a is the y intercept
- b is the ratio between $Δ y$

Derivative

\begin{aligned} a^{x} \\ = e^{\ln a^{x}} \\ = e^{x \ln a} \\ \frac{d}{d x} (e^{x \ln a}) & = \frac{d}{d (x \ln a)} (e^{x \ln a}) \cdot \frac{d}{d x} (x \ln a) \\ = e^{x \ln a} \cdot \ln a \\ = a^{x} \ln a \end{aligned}

Solving

Make both sides have the same bases but different exponents
Just remove the bases
profit

Example

$\begin{aligned} 5^{x} & = 125 \\ 5^{x} & = 5^{3} \\ x & = 3 \end{aligned}$ $\begin{aligned} 7^{x + 3} & = \frac{1}{49} \\ 7^{x + 3} & = 7^{- 2} \\ x + 3 & = - 2 \\ x & = 5 \end{aligned}$ $\begin{aligned} 2^{2 x} - 3 (2)^{x + 2} + 32 & = 0 \\ let a & = 2^{x} \\ then a^{2} & = 2^{2 x} \\ a^{2} - 3 a (4) + 32 & = 0 \\ a^{2} - 12 a + 32 & = 0 \\ (a - 4) (a - 8) & = 0 \\ \begin{aligned} a & = 4 \\ 2^{x} & = 4 \\ 2^{x} & = 2^{2} \\ x & = 2 \end{aligned} & OR \begin{aligned} a & = 8 \\ 2^{x} & = 8 \\ 2^{x} & = 2^{3} \\ x & = 3 \end{aligned} \\ ∴ x & \in {2, 3} \end{aligned}$

Take the log of both sides

Example

$\begin{aligned} 2 (3^{5 x}) & = 7^{x - 2} \\ \log (2 (3^{5 x})) & = \log (7^{x - 2}) \\ \log (2) + 5 x \log (3) & = (x - 2) \log (7) \\ = x \log (7) - 2 \log (7) \\ 5 x \log (3) - x \log (7) & = - 2 \log (7) - \log (2) \\ x (5 \log (3) - \log (7)) & = - 2 \log (7) - \log (2) \\ x & = \frac{- 2 \log (7) - \log (2)}{5 \log (3) - \log (7)} \end{aligned}$