5 — Logarithmic Functions

A structured guide to defining, transforming, graphing, evaluating, and applying logarithmic functions, with careful attention to domains and solution checks.

The Meaning and Notation of Logarithms

A answers the question “What exponent produces this number?” For a valid base $b$ ,

\log_b(x)=y \quad\Longleftrightarrow\quad b^y=x,

where $b>0$ , $b\ne 1$ , and $x>0$ . The condition $x>0$ is essential: the argument of a real cannot be zero or negative.

For example,

\log_2(8)=3

because $2^3=8$ . The common is $\log(x)=\log_{10}(x)$ , and the natural is $\ln(x)=\log_e(x)$ , where $e\approx 2.718$ .

Takeaway: A is an exponent, and its argument must always be positive.

Inverse Relationships

Exponential and logarithmic functions with the same base are . Their defining relationships are

b^{\log_b(x)}=x \quad (x>0)

and

\log_b(b^x)=x.

Thus, a can undo an exponential expression immediately. For example,

\log_5(5^{3x-1})=3x-1.

Because interchange inputs and outputs, their graphs reflect across the line $y=x$ . A point $(2,4)$ on $y=2^x$ corresponds to $(4,2)$ on $y=\log_2(x)$ .

Takeaway: Matching and exponential bases allows the two operations to cancel.

Properties, Expansion, and Condensation

The main properties come from the laws of exponents. For positive $M$ and $N$ ,

Product property: $\log_b(MN)=\log_b(M)+\log_b(N)$
Quotient property: $\log_b\left(\frac{M}{N}\right)=\log_b(M)-\log_b(N)$
Power property: $\log_b(M^p)=p\log_b(M)$
Special values: $\log_b(1)=0$ and $\log_b(b)=1$

To expand an expression, reverse products, quotients, and powers. For example,

\log_2\left(\frac{8x^3}{y}\right)=3+3\log_2(x)-\log_2(y),

provided $x>0$ and $y>0$ . To condense, reverse the process:

2\ln(x)+\ln(y)-\ln(4)=\ln\left(\frac{x^2y}{4}\right).

There is no sum property for addition. In general,

\log_b(M+N)\ne \log_b(M)+\log_b(N).

Takeaway: properties apply to multiplication, division, and powers, not to sums or differences inside an argument.

Changing the Base

The allows a to be evaluated or compared using another valid base:

\log_b(M)=\frac{\log_c(M)}{\log_c(b)}.

The most useful calculator forms are

\log_b(M)=\frac{\ln(M)}{\ln(b)}

and

\log_b(M)=\frac{\log(M)}{\log(b)}.

For example,

\log_5(37)=\frac{\ln(37)}{\ln(5)}\approx 2.240.

The conditions remain important: $M>0$ , $b>0$ , $b\ne 1$ , and the new base must also be positive and different from $1$ .

Takeaway: Convert unfamiliar bases to $\ln$ or common logarithms when evaluating with a calculator.

Graphs and Transformations

The parent function $y=\log_b(x)$ has these features:

Domain: $(0,\infty)$
Range: $(-\infty,\infty)$
: $x=0$
$x$ -intercept: $(1,0)$
No $y$ -intercept
Increasing when $b>1$
Decreasing when $0<b<1$

Useful points are

\left(\frac{1}{b},-1\right),\qquad (1,0),\qquad (b,1).

For a transformed function,

f(x)=a\log_b(x-h)+k,

$h$ shifts the graph horizontally, $k$ shifts it vertically, and the asymptote becomes $x=h$ . The coefficient $a$ controls vertical stretching, compression, and reflection. The domain comes from requiring the argument to be positive. For example, for $f(x)=\log_2(x-3)+1$ , the domain is $x>3$ , the asymptote is $x=3$ , and the graph is shifted right $3$ units and up $1$ unit.

Takeaway: Find the domain and asymptote from the ’s argument before plotting points.

Solving Logarithmic Equations

Several methods solve logarithmic equations. First, convert directly to exponential form when possible:

\log_4(x)=3 \quad\Longleftrightarrow\quad 4^3=x,

so $x=64$ .

When equal logarithms have the same base, use the :

\log_3(2x-1)=\log_3(11)

implies

2x-1=11,

so $x=6$ . The argument is positive at this value, so the solution is valid.

If several logarithms occur, combine them first. Consider

\ln(x)+\ln(x-3)=\ln(10).

The product property gives

\ln\bigl(x(x-3)\bigr)=\ln(10),

x(x-3)=10.

The possible values are $x=5$ and $x=-2$ , but the original logarithms require $x>3$ . Therefore, $x=5$ is valid and $x=-2$ is an .

For an exponential equation with an unknown exponent, take logarithms:

2^x=7 \quad\Longrightarrow\quad x=\frac{\ln(7)}{\ln(2)}\approx 2.807.

In general,

a^x=c \quad\Longrightarrow\quad x=\frac{\ln(c)}{\ln(a)}.

Takeaway: Check every candidate in the original equation, especially after combining logarithms or applying exponentials.

Logarithmic Inequalities

A depends on whether its base is greater than $1$ or between $0$ and $1$ . If $b>1$ , the is increasing and the inequality direction is preserved. If $0<b<1$ , the is decreasing and the direction reverses.

For example, with a base greater than $1$ ,

\log_2(x-1)>3

becomes

x-1>2^3,

so $x>9$ .

With a base between $0$ and $1$ ,

\log_{1/3}(x)\le -2

becomes

x\ge \left(\frac{1}{3}\right)^{-2}=9.

For inequalities containing two logarithms, compare arguments only after checking the domain. For

\log_5(x+2)>\log_5(3x-4),

base $5>1$ preserves the direction, giving $x<3$ . The domain requires $x+2>0$ and $3x-4>0$ , so the complete solution is

x\in\left(\frac{4}{3},3\right).

Takeaway: Apply the base rule and intersect the resulting inequality with every domain condition.

Applications and Exponential Models

Logarithms isolate time or another variable that appears as an exponent. For an exponential growth or decay model,

A(t)=A_0b^t,

solving for time gives

t=\log_b\left(\frac{A(t)}{A_0}\right).

For a continuous model,

A(t)=A_0e^{kt},

solving for time gives

t=\frac{\ln\left(A(t)/A_0\right)}{k}.

For example, suppose

P(t)=500(1.08)^t

and the target population is $750$ . Then

750=500(1.08)^t,

1.5=(1.08)^t.

Taking natural logarithms produces

t=\frac{\ln(1.5)}{\ln(1.08)}\approx 5.27.

The target is reached after approximately $5.27$ years.

Takeaway: When the unknown is an exponent, logarithms convert the exponential relationship into a solvable quotient.