Section A.1 Exponential and Logarithmic Functions

Recall the following rules for exponential and logarithmic functions.

Exponential Rules.

	General	Natural ( $a = e$ )
$E_{1} :$	$\begin{matrix} (a b)^{x} = a^{x} \cdot b^{x} \end{matrix}$	$\begin{matrix} (e b)^{x} = e^{x} \cdot b^{x} \end{matrix}$
$E_{2} :$	$\begin{matrix} a^{x} \cdot a^{y} = a^{x + y} \end{matrix}$	$\begin{matrix} e^{x} \cdot e^{y} = e^{x + y} \end{matrix}$
$E_{3} :$	$\begin{matrix} (a^{x})^{y} = a^{x y} \end{matrix}$	$\begin{matrix} (e^{x})^{y} = e^{x y} \end{matrix}$
$E_{4} :$	$\begin{matrix} a^{- x} = \frac{1}{a^{x}} \end{matrix}$	$\begin{matrix} e^{- x} = \frac{1}{e^{x}} \end{matrix}$

Logarithmic Rules.

	General	Natural ( $a = e$ )
$L_{1} :$	$\begin{matrix} \log_{b} (b^{x}) = x \end{matrix}$	$\begin{matrix} \ln (e^{x}) = x \end{matrix}$
$L_{2} :$	$\begin{matrix} b^{\log_{b} (x)} = x \end{matrix}$	$\begin{matrix} e^{\ln (x)} = x \end{matrix}$
$L_{3} :$	$\begin{matrix} \log_{b} (1) = 0 \end{matrix}$	$\begin{matrix} \ln (1) = 0 \end{matrix}$
$L_{4} :$	$\begin{matrix} \log_{b} (x y) = \log_{b} (x) + \log_{b} (y) \end{matrix}$	$\begin{matrix} \ln (x y) = \ln (x) + \ln (y) \end{matrix}$
$L_{5} :$	$\begin{matrix} \log_{b} (\frac{x}{y}) = \log_{b} (x) - \log_{b} (y) \end{matrix}$	$\begin{matrix} \ln (\frac{x}{y}) = \ln (x) - \ln (y) \end{matrix}$

Let’s look at a couple of examples, starting with an equation containing exponentials.

Example A.1.

Solve for

x :

e^{3 x + z} - e^{z} = y

Solution. Solution

If you prefer to see a video of this example, click here

⁹

www.youtube.com/watch?v=QdZekQ5_T2E

We might begin by isolating the exponential that contains

x

and then taking the natural log of both sides.

\begin{aligned} e^{3 x + z} - e^{z} = & y \\ e^{3 x + z} = & y + e^{z} \\ \overset{L_{1}}{↪} \ln (e^{3 x + z}) = & \ln (y + e^{z}) \\ 3 x + z = & \ln (y + e^{z}) \\ 3 x = & \ln (y + e^{z}) - z \\ x = & \frac{1}{3} \ln (y + e^{z}) - \frac{1}{3} z \end{aligned}

It’s worth noting that we cannot break up that log on the right hand side. There’s no "rule" that helps when we have addition inside a logarithm.

There is another way to approach this if notice that

z

appears inside both exponential terms.

\begin{aligned} \overset{E_{2}}{↪} e^{3 x + z} - e^{z} = & y \\ e^{3 x} \cdot e^{z} - e^{z} = & y \\ e^{z} (e^{3 x} - 1) = & y \\ e^{3 x} - 1 = & \frac{y}{e^{z}} \overset{E_{4}}{↩} \\ e^{3 x} - 1 = & y e^{- z} \\ e^{3 x} = & y e^{- z} + 1 \\ \overset{L_{1}}{↪} \ln (e^{3 x}) = & \ln (y e^{- z} + 1) \\ 3 x = & \ln (y e^{- z} + 1) \\ x = & \frac{1}{3} \ln (y e^{- z} + 1) \end{aligned}

The answers may look different, but they are equivalent and both are correct.

Now let’s look at an example involving logarithms.

Example A.2.

Solve for

x :

\ln (x + y) = 5 + \ln (z)

Solution. Solution

If you prefer to see a video of this example, click here

¹⁰

www.youtube.com/watch?v=7xG6SisdjYE

We’ll carefully apply the rules above. We want to get our hands on

x,

and right now its inside a logarithm. In order to undo that, we’ll exponentiate both sides.

\begin{aligned} \ln (x + y) = & 5 + \ln (z) \\ \overset{L_{2}}{↪} e^{\ln (x + y)} = & e^{(5 + \ln (z))} \overset{E_{2}}{↩} \\ x + y = & e^{5} \cdot e^{\ln (z)} \overset{L_{2}}{↩} \\ x = & e^{5} z - y \end{aligned}

Now you should try. Be careful!

Use algebra and the rules above to solve for

x

in each of the following equations.

$e^{x + y} = 12$
Solution. Solution
$\begin{aligned} e^{x + y} = & 12 \\ \ln (e^{x + y}) = & \ln (12) \\ x + y = & \ln (12) \\ x = & \ln (12) - y \end{aligned}$

Answer. Answer
$x = \ln (12) - y$
$e^{x + y} + e^{x} = 12$
Solution. Solution
$\begin{aligned} e^{x + y} + e^{x} = & 12 \\ e^{x} \cdot e^{y} + e^{x} = & 12 \\ e^{x} (e^{y} + 1) = & 12 \\ e^{x} = & \frac{12}{e^{y} + 1} \\ \ln (e^{x}) = & \ln (\frac{12}{e^{y} + 1}) \\ x = & \ln (\frac{12}{e^{y} + 1}), or \\ = & \ln (12) - \ln (e^{y} + 1) \end{aligned}$

Answer. Answer
$\begin{aligned} x = & \ln (\frac{12}{e^{y} + 1}), or \\ = & \ln (12) - \ln (e^{y} + 1) \end{aligned}$
$e^{x} = 1$
Solution. Solution
$\begin{aligned} e^{x} = & 1 \\ \ln (e^{x}) = & \ln (1) \\ x = & 0 \end{aligned}$

Answer. Answer
$x = 0$
$\ln x = 3 \ln z$
Solution. Solution
$\begin{aligned} \ln x = & 3 \ln z \\ e^{\ln x} = & e^{3 \ln z} \\ e^{\ln x} = & e^{\ln (z^{3})} \\ x = & z^{3} \end{aligned}$

Answer. Answer
$x = z^{3}$
$y + \ln x = 4$
Solution. Solution
$\begin{aligned} y + \ln x = & 4 \\ \ln x = & 4 - y \\ e^{\ln x} = & e^{4 - y} \\ x = & e^{4 - y} \end{aligned}$

Answer. Answer
$x = e^{4 - y}$
$\ln y + \ln x = 4$
Solution. Solution
$\begin{aligned} \ln y + \ln x = & 4 \\ \ln x = & 4 - \ln y \\ e^{\ln x} = & e^{4 - \ln y} \\ x = & e^{4 - \ln y} \\ = & e^{4} \cdot e^{- \ln y} \\ = & e^{4} \cdot e^{\ln (y^{- 1})} \\ = & e^{4} \cdot y^{- 1} \\ = & e^{4} \cdot \frac{1}{y} \\ = & \frac{e^{4}}{y} \end{aligned}$

Answer. Answer
$x = \frac{e^{4}}{y}$
$\ln x = 8 \ln y + 5$
Solution. Solution
$\begin{aligned} \ln x = & 8 \ln y + 5 \\ e^{\ln x} = & e^{8 \ln y + 5} \\ x = & e^{8 \ln y} \cdot e^{5} \\ = & e^{\ln (y^{8})} \cdot e^{5} \\ = & y^{8} \cdot e^{5} \\ = & e^{5} y^{8} \end{aligned}$

Answer. Answer
$x = e^{5} y^{8}$

Definition A.3. Euler’s Formula.

e^{i θ} = \cos (θ) + i \sin (θ)

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