Common log (log₁₀) and natural log (ln x): Worked examples

The one tool we lean on: change of base

Before the examples, let us state the bridge we use whenever a number is not a clean power of the base — and, because this page is visual-first, let us actually see why it works.

Where does it come from? Let $y = lo g_{b} a$ — that is, $y$ is the unknown power with $b^{y} = a$ . Now take $ln$ of both sides: $ln (b^{y}) = ln a \Rightarrow y ln b = ln a \Rightarrow y = \frac{l n a}{l n b} .$ Why this step? The unknown $y$ is stuck in an exponent; $ln$ pulls it down via the power law $ln (b^{y}) = y ln b$ , turning a hard "hidden-exponent" equation into a one-line division.

The scenario matrix

We are always answering " $b$ to what power $= x$ ?" — so the two knobs are: what is $x$ and what is the base $b$ . Here is every distinct behaviour:

Cell	Situation	What makes it special	Example
A	$x$ is an exact power of the base ( $x = b^{k}$ , $k$ whole)	Answer is a clean integer — no calculator	Ex 1
B	$x$ between $0$ and $1$	Answer is negative (fewer than one "whole" multiply)	Ex 2
C	$x$ is an awkward number, not a power (either base)	Need change-of-base / calculator	Ex 3
D	Unknown sits in the exponent ( $e^{k x} = c$ )	Take $ln$ to pull it down	Ex 4
E	Base-10 equation solved via $ln$	Change-of-base appears automatically	Ex 5
F	$x = 1$ , and $x \to 0^{+}$ , $x \to \infty$ (limits)	Degenerate & boundary behaviour	Ex 6
G	$x \leq 0$ (zero / negative)	Undefined — must say so	Ex 7
H	Real-world word problem (growth/decay, pH)	Set up equation from words	Ex 8
I	Exam twist — log inside log / combining laws	Careful sequencing	Ex 9

The graph below shows why the sign of the answer flips as $x$ crosses $1$ — the single most important shape to hold in your head.

Cell A — exact power of the base

Cell B — number between 0 and 1

Cell C — awkward number, not a clean power (works the same in any base)

Worked example Example 3a —

$lo g_{10} (50)$ Evaluate to 4 decimal places, and check using digit-counting intuition.

Forecast: $50$ lies between $1 0^{1} = 10$ and $1 0^{2} = 100$ , so $lo g_{10} (50)$ must be between $1$ and $2$ . Closer to which? $50$ is past halfway to $100$ , so expect roughly $1.7$ .

There is no whole power of $10$ equal to $50$ , so we cannot read the answer off. Why this step matters: recognising "not a clean power" tells us we need the change-of-base tool, not mental arithmetic.
Apply the change-of-base formula $lo g_{b} a = \frac{ln a}{ln b}$ with $a = 50$ , $b = 10$ : $lo g_{10} (50) = \frac{ln 50}{ln 10}$ . Why? The formula converts to logs the calculator actually has.
$\frac{ln 50}{ln 10} = \frac{3.9120}{2.3026} = 1.6990$ . Why divide? That is precisely what change-of-base says: wanted-number's $ln$ over base's $ln$ .

Verify: $1 0^{1.6990} \approx 50.0$ ✓, and $1.6990$ sits between $1$ and $2$ as forecast ✓.

Worked example Example 3b —

$ln (17)$ (same trick, base $e$ ) Evaluate $ln 17$ to 4 decimal places using only the mental anchors $e \approx 2.718$ , $e^{2} \approx 7.39$ , $e^{3} \approx 20.1$ .

Forecast: $17$ lies between $e^{2} \approx 7.4$ and $e^{3} \approx 20.1$ , so $ln 17$ must be between $2$ and $3$ . Since $17$ is quite near $20$ , expect something like $2.8$ .

$17$ is not a clean power of $e$ , so we cannot read the answer off directly. Why this step matters: the awkward-number situation is identical whatever the base — the base being $e$ instead of $10$ changes nothing about the method.
Here the base is already $e$ , so no conversion is even needed — the calculator's ln button is $lo g_{e}$ : $ln 17 = 2.8332$ . Why point this out? It shows change-of-base is the general rule and base 10 was never special; base $e$ is just the case where the ruler and the tool already match.
Sanity via change-of-base the other way: $ln 17 = lo g_{10} 17 \cdot ln 10 = 1.2304 \times 2.3026 = 2.8332$ . Why? Confirms both rulers give the same underlying "how many multiplies," differing only by the factor $ln 10$ .

Verify: $e^{2.8332} \approx 17.0$ ✓, and $2.8332$ sits between $2$ and $3$ as forecast ✓.

Cell D — unknown in the exponent (natural base)

Cell E — base-10 equation via natural log

Cell F — the boundary $x = 1$ and the limits

Worked example Example 6 —

$ln 1$ , and the behaviour as $x \to 0^{+}$ and $x \to \infty$ Evaluate $ln 1$ ; describe what $ln x$ and $lo g_{10} x$ do at the two edges of their (shared) domain.

Forecast: Multiplying the base zero times leaves you at $1$ , so $ln 1$ (and $lo g_{10} 1$ ) should be $0$ . As $x$ shrinks toward $0$ the log should plunge; as $x$ grows the log should climb — but slowly.

$ln 1 = 0$ , and likewise $lo g_{10} 1 = 0$ . Why? $e^{0} = 1$ and $1 0^{0} = 1$ , so "base to what gives 1?" answers $0$ for any base: $lo g_{b} 1 = 0$ .
As $x \to 0^{+}$ : $ln x \to - \infty$ , and $lo g_{10} x \to - \infty$ too. Why? To reach a tiny positive number you need the base raised to a very negative power — true whether the base is $e$ or $10$ . In the graph above the red curve dives down without bound near the vertical axis.
As $x \to \infty$ : both $ln x \to + \infty$ and $lo g_{10} x \to + \infty$ , but slowly. Why? Each extra unit of log requires multiplying $x$ by the base again — so $x$ has to grow explosively for the log to inch up. This boundary behaviour is not base-specific; the two curves are just scaled copies of each other (they differ only by the factor $ln 10$ ).

Verify: $e^{0} = 1$ and $1 0^{0} = 1$ ✓. Numerically $ln (0.001) \approx - 6.9$ and $lo g_{10} (0.001) = - 3$ (both heading to $- \infty$ ); $ln (1000) \approx 6.9$ and $lo g_{10} (1000) = 3$ (both still small vs 1000) — confirms the slow-climb, steep-dive shape for both bases.

Cell G — zero and negative inputs (the forbidden zone)

Worked example Example 7 — Evaluate

$lo g_{10} (0)$ and $ln (- 5)$ State the answers.

Forecast: No power of $10$ (or $e$ ) can be zero or negative — powers of a positive base are always positive. So expect "undefined."

$lo g_{10} (0)$ : we need $1 0^{y} = 0$ . Why undefined? $1 0^{y}$ gets arbitrarily close to $0$ as $y \to - \infty$ but never equals $0$ . No finite $y$ works, so $lo g_{10} 0$ is undefined (it "tends to $- \infty$ ").
$ln (- 5)$ : we need $e^{y} = - 5$ . Why undefined? $e^{y} > 0$ for every real $y$ — the exponential curve lives entirely above the axis. There is no real $y$ giving a negative result.
Domain rule: both $lo g_{10} x$ and $ln x$ require $x > 0$ . Any zero or negative input is outside the domain in the reals.

Verify: $1 0^{y} > 0$ and $e^{y} > 0$ for all real $y$ — in the graph above the red curve exists only to the right of $x = 0$ ; there is literally no curve above $x \leq 0$ ✓.

Cell H — real-world word problem

Worked example Example 8 — Radioactive decay half-life

A sample decays by the model $N (t) = N_{0} e^{- 0.05 t}$ ( $t$ in years). How long until only half remains?

Forecast: Slow decay ( $0.05$ per year is small), so the half-life is much longer than a year — likely somewhere around 10–15 years.

"Half remains" means $N (t) = \frac{1}{2} N_{0}$ , so $\frac{1}{2} N_{0} = N_{0} e^{- 0.05 t}$ . Why this step? Translate the words "half remains" into an equation.
Divide by $N_{0}$ : $\frac{1}{2} = e^{- 0.05 t}$ . Why? The starting amount cancels — half-life doesn't depend on how much you started with.
Take $ln$ : $ln (\frac{1}{2}) = - 0.05 t$ . Why $ln$ and not $lo g_{10}$ ? The base is $e$ , so $ln$ undoes it in one clean step (no stray $ln 10$ factor).
$t = \frac{ln ( 1/2 )}{- 0.05} = \frac{- 0.6931}{- 0.05} = 13.86$ years. Why divide by $- 0.05$ ? Isolate $t$ ; the two negatives cancel to a positive time (good — time can't be negative).

Verify: $N_{0} e^{- 0.05 \times 13.86} = N_{0} e^{- 0.6931} = N_{0} \times 0.5$ ✓. Units: years ✓. Forecast (10–15 yrs) ✓. See Exponential growth and decay models.

Cell I — exam twist: laws combined with a log

Scenario coverage check

Recall Did we hit every cell?

A (exact power) — Ex 1 ✓ · B (0 to 1) — Ex 2 ✓ · C (awkward, both bases) — Ex 3a & 3b ✓ · D (natural-base exponent) — Ex 4 ✓ · E (base-10 via ln) — Ex 5 ✓ · F ( $x = 1$ & limits) — Ex 6 ✓ · G (zero/negative) — Ex 7 ✓ · H (word problem) — Ex 8 ✓ · I (exam twist) — Ex 9 ✓. Every cell of the matrix has a worked example.

Recall

Which cell needs change-of-base?

Cell C — an awkward

x

that is not a clean power of the base.

Why is

lo g_{10} (0.01)

negative?

Because

0.01 = 1 0^{- 2}

; numbers below 1 need negative powers.

What frees an unknown stuck in an exponent?

Taking

ln

(or

lo g

) of both sides — logs pull exponents down.

Why is

ln (- 5)

undefined?

e^{y} > 0

for all real

y

, so no power of

e

gives a negative number.

After solving a log equation, what must you always do?

Check each root keeps every log's argument

> 0

(domain check).

Half-life of

N_{0} e^{- 0.05 t}

t = \frac{ln ( 1/2 )}{- 0.05} \approx 13.86

years.

Solve

lo g_{10} x + lo g_{10} (x - 3) = 1

x = 5

(reject

x = - 2

Evaluate

ln 17

\approx 2.8332

(between

2

and

3

, since

e^{2} < 17 < e^{3}

Connections

Laws of Logarithms (product, quotient, power) — the product law powers Example 9.
Solving exponential equations — cells D and E live here.
Exponential functions e^x and 10^x — why powers of positive bases stay positive (cell G).
Exponential growth and decay models — home of Example 8.
Scientific notation and orders of magnitude — why cells A and B are digit-counting.
Differentiation of ln x and e^x — the calculus behind the slow-climb shape in cell F.

Concept Map

The whole topic is one question with a single gate followed by a menu of techniques. Every logarithm starts by asking "the base to what power gives $x$ ?" The first fork is the domain gate: is $x > 0$ ? If not, the road ends — the log is undefined, because no power of a positive base is zero or negative (cell G). If yes, the kind of answer depends on $x$ : an exact power gives a clean integer (cell A); an $x$ between 0 and 1 gives a negative answer (cell B); an awkward $x$ (in any base) needs the change-of-base tool (cell C); an $x$ trapped in an exponent is freed by taking a log (cell D); and at $x = 1$ the answer is exactly zero (cell F). Read the diagram below as that gate-then-menu flow.