Logarithmic Rules

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rules

Exponents

Logarithms are exponents.

$\log _a(b)$ is what you raise $a$ to, to get $b$ .

$a^{\log _a(b)} = b \, \text { for } \, a,b > 0$

We can read it right to left as well.

Any positive number, $b$ , can be written with any base, $a$ , like $b = a^{\log _a(b)}$ .

Logarithms are exponents. Therefore, they should follow all of the exponent rules.

$\blacktriangleright$ Let $M$ and $N$ be two positive real numbers.

We can write them as $M = a^{\log _a(M)}$ and $N = a^{\log _a(N)}$

This allows us to write the product $M \cdot N$ in two different ways.

$M \cdot N = a^{\log _a(M)} \cdot a^{\log _a(N)}$

$M \cdot N = a^{\log _a(M \cdot N)}$

Therefore, these must be equal.

$a^{\log _a(M)} \cdot a^{\log _a(N)} = a^{\log _a(M \cdot N)}$

Apply an exponent rule:

$a^{\log _a(M)+\log _a(N)} = a^{\log _a(M \cdot N)}$

Since exponential functions are one-to-one, we have

$\log _a(M)+\log _a(N) = \log _a(M \cdot N)$

$\log _a(M)+\log _a(N) = \log _a(M \cdot N) \, \text { for } \, a, M, N > 0$

$\blacktriangleright$ Let’s write the quotient $\frac {M}{N}$ in two different ways.

$\frac {M}{N} = \frac {a^{\log _a(M)}}{a^{\log _a(N)}}$

$\frac {M}{N} = a^{\log _a\left (\frac {M}{N}\right )}$

Therefore, these must be equal.

$\frac {a^{\log _a(M)}}{a^{\log _a(N)}} = a^{\log _a\left (\frac {M}{N}\right )}$

Apply an exponent rule:

$a^{\log _a(M) - \log _a(N)} = a^{\log _a\left (\frac {M}{N}\right )}$

Since exponential functions are one-to-one, we have

$\log _a(M)-\log _a(N) = \log _a\left (\frac {M}{N}\right )$

$\log _a(M)-\log _a(N) = \log _a\left (\frac {M}{N}\right ) \, \text { for } \, a, M, N > 0$

$\blacktriangleright$ Let’s write $M^N$ in two different ways.

$M^N = a^{\log _a(M^N)}$

$M^N = (a^{\log _a(M)})^N = (a^{N \cdot \log _a(M)})$

Therefore, these must be equal.

$a^{\log _a(M^N)} = (a^{N \cdot \log _a(M)})$

Since exponential functions are one-to-one, we have

$\log _a(M^N) = N \cdot \log _a(M)$

$\log _a(M^N) = N \cdot \log _a(M) \, \text { for } \, a, M, N > 0$

One-to-One

Since exponential functions are one-to-one, and logarithmic functions are just the reverse, logarithmic functions must be one-to-one as well. One-to-one means that each range number is paired with a unique domain number.

In other words, each function value in a basic logarithmic function occurs exactly once.

If you know that $\log _a(r) = \log _a(t)$ , then $r=t$ follows.

This is an important rule.

Uniqueness

$\text {If } \, \log _a(r) = \log _a(t), \, \text { then } \, r=t$

Solving Equations

Solve $\log _5(21 + x) = 2$

$\log _5(21 + x) = 2$

$5^{\log _5(21 + x)} = 5^2$

$21 + x = 5^2 = 25$ , using the definition of logarithm.

$x = 4$

Solving Equations

Solve $\log _2(y) + \log _2(y-4) = \log _2(y+6)$

$\log _2(y) + \log _2(y-4) = \log _2(y+6)$

$\log _2(y(y-4)) = \log _2(y+6)$

$y(y-4) = y+6$

$y^2 - 4y - y - 6 = 0$

$y^2 - 5y - 6 = 0$

$(y-6) \left ( \answer {y+1} \right ) = 0$

Either $y-6 = 0$ or $y+1 = 0$

Either $y = 6$ or $y = -1$

However, $y \ne -1$ , since $\answer {-1}$ is not in the domain of $\log _2(y)$ .

Therefore, just one solution: $6$ .

The solution set is $\{ 6 \}$

Solving Equations

Solve $7^k = 5$

We are looking for the number that you raise $7$ to, to get $5$ . That number is $\log _7(5)$ .

Change of Base

Logarithms are exponents. We use them to write expressions in exponential form.

Any positive number, $b$ , can be written as a power of another positive number, $a$ .

$b = a^{\log _a(b)}$

For this to be useful, we will need to write multiple expressions with the same base. Therefore, changing the base becomes important.

How do we write $\log _a(b)$ in terms of $\log _c(x)$ ?

First, $a$ and $b$ can be written in terms of some third base, $c$ , using logarithms.

$b = c^{\log _c(b)} \, \text { and } \, a = c^{\log _c(a)}$

Second, substitute these in for the $a$ and $b$ bases in $b = a^{\log _a(b)}$ .

$c^{\log _c(b)} = \left (c^{\log _c(a)}\right )^{\log _a(b)}$

$c^{\log _c(b)} = c^{\log _c(a) \cdot \log _a(b)}$

$\log _c(b) = \log _c(a) \cdot \log _a(b)$

$\frac {\log _c(b)}{\log _c(a)} = \log _a(b)$

This is known as the Change of Base Formula.

Change of Base Formula

$\log _a(b) = \frac {\log _c(b)}{\log _c(a)} \, \text { for } \, a, b, c > 0$

Write $\log _3(5)$ in terms of logarithms base $7$ .

Following the change of base template: $\log _a(b) = \frac {\log _c(b)}{\log _c(a)}$ gives

$\log _3(5) = \frac {\log _7(5)}{\log _7(3)}$

$\blacktriangleright$ It occured to people: If we can write any power in terms of any base, and any logarithm in terms of any base, then let’s just pick one base to write everything in.

We have such a base: $e$

e

$e$ was defined in a very weird way.

$e = \lim _{x \to \infty } \left (\answer {1 + \frac {1}{x}}\right )^x$

It is weird now. But as you continue through Calculus, you will see $e$ pop up all over the place. It seems to have a connection to everything. So much so that scientists, engineers, and mathematicians have adopted $e$ as the prefered base for everything.

This means that we encounter $\log _e(x)$ a lot. And, any time something appears a lot in mathematics, it usually gets a shortcut abbreviation.

Natural Logarithm

$\ln (x) = \log _e(x)$

That is a lowerscase “el” and a lowercase “en”.

That is not an uppercase “eye”. It is not In.

It is a lowercase “el”, ln.

base e

$\log _7(5) = \frac {\ln (5)}{\ln (7)}$

Everything can be written in terms of base $e$ .

Calculators usually have a button titled “ln” or “LN”. When approximating values of logarithms with other bases, we convert them to natural logarithms. Then we can use the calculator.

Approximate $\log _{13}(61)$

Using the LN button on the calculator, we get $\log _{13}(61) = \frac {\ln \left (\answer {61}\right )}{\ln \left (\answer {13}\right )} \approx 1.602711512$

Rewrite the expression $3 \cdot 5^{2x + 1}$ in the form $e^{f(x)}$ .

$3 = e^{\answer {\ln (3)}}$

$5 = e^{\answer {\ln (5)}}$

$3 \cdot 5^{2x + 1} = e^{\ln (3)} \cdot (e^{\ln (5)})^{2x + 1} = e^{\ln (3)} \cdot e^{\ln (5) \cdot (2x + 1)} = e^{\ln (3) + \ln (5) \cdot (2x + 1)}$

$e^{\ln (3) + \ln (5) \cdot (2x + 1)}$

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