# Unit 2: Exponential and Logarithmic Functions


:::{.proposition title="Multiplicative and additive properties of logs and exponentials"}
For logarithms:

1. $\log_\alpha(a) + \log_\alpha(b) = \log_\alpha(ab)$
2. $\log_\alpha(b^a) = a\log_{\alpha}(b)$
  Mnemonic: follows from (1) for $a\in \NN$ since exponentiation is repeated multiplication.
3. $\log_\alpha(a + b) = \text{ nothing!}$

For exponentials:

1. $(\alpha^a)^b = \alpha^{ab}$
2. $\alpha^a \alpha^b = \alpha^{a+b}$
3. $\alpha^a + \alpha^b = \text{ nothing!}$

:::


:::{.proposition title="Inverse properties"}
The pair $f(x) = \alpha^x$ and $g(x) = \log_\alpha(x)$ form an inverse pair.
Thus
\[
(f\circ g)(?) &= \alpha^{\log_\alpha(?)} = ? \\
(g\circ f)(?) &= \log_\alpha(\alpha^{?}) = ?
.\]

Moreover, both functions are **injective**, so
\[
a &= b \\
\iff \alpha^a &= \alpha^b \\
\iff \log_\alpha(a) &= \log_\alpha(b)
.\]

:::


:::{.example title="?"}
Simplifying:

- $e^{5\ln(x)} = x^5$.
- $\ln(e^{5x}) = 5x$
- $10^{3 \log_{10}(x) - 4 \log_{10}(x^2 + 1)} = x^3 / (x^2 + 1)^4$.


Solving for $x$:

- In $\log_\alpha(x+1) + \log_\alpha(x-1) = \beta$, can obtain $x^2 = \alpha^\beta + 1$.

:::

## Algebra


:::{.example title="Simplifying"}

- $\qty{y^2 \over x}^3 {x^4 \over y} = y^5 x$
- $(-2x^2 t^3 (z+\ell)^2 )^{-2} {z^3\over x} = -{1\over 2} x^{-5} (z+\ell)^{-4} z^3 t^{-6}$.
- $3^x = {1\over 5}7^x$ implies $x = \ln(1/5) / (\ln(3) - \ln(7))$.
  - Common mistake: $\ln({1\over 5} 7^x) \neq x\ln( {1\over 5} 7)$.
  Need exponent to be on the *entire* argument.

:::