Logarithms and Exponents

I feel like this area needed some review, so here are some notes on various things. Logarithmic functions are inverse of exponential functions, so:

\[ y = log_a x \iff x = a^y \]

This page begins with some notes on videos of NancyPi, then Professor Leonard. The latter began from a more basic starting point of exponential functions.

Nancy Pi Notes

Easy examples

$ log_3\space9 $, think to yourself 3 to what power equals 9, but formally, do this, make it equal to x, i.e. $ x = log_3\space 9 $. Next use formula above so $ x = log_3\space 9 \iff 3^x = 9 $ therefore x = 2, or $ log_3\space 9 = 2 $. With no base, e.g. $ log 10000 $, then base = 10, so equals $log_{10} \space10000 = x \iff x^10 = 10000 = 4$.

Weirder logs

$ log_2\space\frac{1}{8} = x \iff 2^x = \frac{1}{8}.\space \therefore \space x = -3$

Other weird ones:

$ log\space 1 (1); log\space0 (undefined); log\space(-1) (undefined)$

Natural logs

$ ln\space1 = log_e\space1 $

Do same steps as before, set it to x and rearange to exponential form:

$ log_e\space1 = x \iff x^x = 1 \therefore x = 0 $

$ ln(e^3) $ Visually, this should be 3! Do it long way:

$ ln_e (e^3) = x \iff e^x = e^3 \therefore x = 3 $

Even weirder

\[ \begin{align}\begin{aligned} log_x\space32 = 5 \iff x^5 = 32 x = 2$ $\\ $$ log_5x = 3 \iff 5^3 = x; x = 125\end{aligned}\end{align} \]

\[ log_2\space7 \iff 2^x = 7 \]

Change of base formula says we can do:

\[ \frac{log\space7}{log\space2} \]

You need to plug that into a calculator. Answer by the way is 2.807354922057604.

Log Properties

$ log_a(x * y ) = log_a\space x + log_a\space y $

$ log_a\space(\frac{x}{y} ) = log_a\space x - log_a\space y $

$ log_a\space x^n = n\space log_a\space x $

More solving

E.g.

\[ \log\space (12x-7) = \log\space (3x+11) \]

Looks hard but if logs have same base, can just set insides = to each other and solve, so

$ 12x -7 = 3x + 11 \\ 9x = 18 \\ x = 2 $

Based on equality property:

\[ log_b \space m = log_b \space m \iff m = n \]

Note, need to plug answer back into original equation – occasionally doesn’t work. Above we get $ log 17 = log 17 $. If get negative number, have to throw it out because log of negative # is undefined.

Now, based on product property:

$ log_a(x * y ) = log_a\space x + log_a\space y $

if have this problem:

$ \log_2 x + \log_2 (x-2) = 3 $

Equivalent to

\[\begin{split} \log_2 \space (x * (x -2)) \\ \log_2 \space \therefore \\ 2^3 = x(x-2) \\ 8 = x^2 - 2x \\ x^2 - 2x - 8 = 0 \\ (x -4)(x + 2) = 0 \\ x = 4, x = -2 \\ x = 4, \cancel{(x = 2)} \end{split}\]

Have to plug back in… which leads us to throw out the -2 as it’s a domain error.

Professor Leonard

Graphs of Exponential Functions

Polynomials have variables, sometimes raised to a constant exponent. E.g. $ax^2 + bx + c = 0$ is a polynomial. In contrast an exponential function has the variable as the exponent, e.g. $f(x) = a^x$, where a is always positive, not 0 or 1. As base gets > 1, get an increasing graph. As it gets less than one (but not negative), get decreasing graph. E.g. this graph. All this assumes that exponents positive too. Remember, negative exponents create reciprocals, e.g. $2^{-1} = \frac{1}{2}$.

Exponential functions have a key point at (0, 1) and a horizontal asymptote. Domain is all reals, but range is always positive.