Logarithm — definition as inverse of exponential: Worked examples

Before anything else, one habit. Whenever you see $lo g_{b} y$ , silently translate it into a question:

The scenario matrix

Here is the full landscape of cases a logarithm problem can live in. Every cell is covered by an example further down.

#	Case class	What makes it distinct	Covered by
A	Answer is a positive whole number	$y > 1$ , base $> 1$ ⇒ exponent positive	Example 1
B	Answer is zero	input $y = 1$ exactly — the pivot point	Example 2
C	Answer is negative	input $0 < y < 1$ , so exponent dips below $0$	Example 3
D	Answer is a fraction	$y$ is a root of the base	Example 4
E	Base between 0 and 1 (decreasing)	signs flip vs. the usual intuition	Example 5
F	Degenerate input	$y \leq 0$ , or the edges $y \to 0^{+}$ and $y \to \infty$	Example 6
G	Degenerate base	$b \leq 0$ or $b = 1$ — no log exists at all	Example 6b
H	Cancellation identities in disguise	nested $b^{l o g}$ / $lo g (b^{\dots})$	Example 7
I	Real-world word problem	pick the base, translate the story	Example 8
J	Exam twist — solve for the base	the base is the unknown, not the exponent	Example 9

We will draw one picture that shows cases A–C and F together on the same graph, because they are all "read a height off the exponential curve and report the exponent" — they only differ in where on the curve you land.

Figure — Logarithm — definition as inverse of exponential

Look at the curve above (it is $y = 2^{x}$ ). Each labelled dot is one of the inputs we are about to read: the orange dot labelled "Case A" is a point above height $1$ (positive exponent); the plum dot labelled "Case B" sits at height $1$ (exponent $0$ ); the brown dot labelled "Case C" is below height $1$ but still above the dashed floor (negative exponent); and the black ✕ labelled "Case F" sits on the line $y = 0$ that the curve only approaches and never reaches — so no dot of the curve lands there and the log is undefined. The log just reads the horizontal position of each point on the curve.

Because a log is the inverse of the exponential, we can also draw the log curve directly. The next figure plots $y = lo g_{2} x$ — the reflection of $y = 2^{x}$ in the line $y = x$ — so you can see the answers of every case as a single smooth curve.

On this log curve the horizontal axis is now the input $x$ and the vertical axis is the answer $lo g_{2} x$ . Notice both ends: as $x$ slides toward $0$ from the right the curve plunges down to $- \infty$ (the vertical asymptote $x = 0$ , matching Case F's shrinking input), and as $x$ grows without bound the curve keeps rising toward $+ \infty$ — slowly, but forever. Those two "ends" are the complete story of where log values can go.

The worked examples

Worked example Example 1 — Case A: positive whole-number answer

Evaluate $lo g_{2} 16$ .

Forecast: guess the answer before reading on. $16$ is bigger than $1$ , and the base $2$ is bigger than $1$ , so the exponent should be a positive number. How many times do you double from $1$ to reach $16$ ?

Translate to a question. $lo g_{2} 16 = ?$ means " $2$ to what power is $16$ ?" Why this step? The definition $lo g_{b} y = x ⟺ b^{x} = y$ tells us the value is literally the answer to that question.
List the powers of the base until we hit the target. $2^{1} = 2, 2^{2} = 4, 2^{3} = 8, 2^{4} = 16$ . Why this step? When the target is an exact power of the base, we can match it by inspection — no fancy machinery needed.
Read off the exponent. We reached $16$ at exponent $4$ , so $lo g_{2} 16 = 4$ . Why this step? Because $lo g_{2} 16$ was defined as that exponent — once the matching power is found, the log's value is that exponent, nothing more to compute.

$lo g_{2} 16 = 4$

Verify: put it back into exponential form: $2^{4} = 16$ . ✓ On the figure this is the orange "Case A" dot — its horizontal position is $4$ .

Worked example Example 3 — Case C: a negative answer

Evaluate $lo g_{2} \frac{1}{8}$ .

Forecast: $\frac{1}{8}$ is less than $1$ , but doubling from $1$ only ever makes things bigger. So to get below $1$ you must go the other way — a negative exponent. Predict the sign, then find the size.

Translate. $lo g_{2} \frac{1}{8} = ?$ means " $2$ to what power is $\frac{1}{8}$ ?" Why this step? Reflex.
Rewrite the target as a power of the base. $\frac{1}{8} = \frac{1}{2 ^{3}}$ , and a reciprocal is a negative exponent: $\frac{1}{2 ^{3}} = 2^{- 3}$ . Why this step? The rule $b^{- n} = \frac{1}{b ^{n}}$ is exactly how the exponential produces numbers below $1$ . Recognising it turns the fraction into a clean power.
Read off. The exponent is $- 3$ , so $lo g_{2} \frac{1}{8} = - 3$ . Why this step? The definition says the log's value equals the matching exponent; step 2 matched $2^{- 3} = \frac{1}{8}$ , so that exponent $- 3$ is the answer.

$lo g_{2} \frac{1}{8} = - 3$

Verify: $2^{- 3} = \frac{1}{8}$ . ✓ On the s01 figure this is the brown "Case C" dot — below height $1$ but above the dashed floor, at horizontal position $- 3$ , to the left of centre.

Worked example Example 4 — Case D: a fractional answer

Evaluate $lo g_{9} 3$ .

Forecast: $3$ is smaller than the base $9$ but still bigger than $1$ . So the exponent is positive, yet less than $1$ — a fraction. Which fraction turns $9$ into $3$ ?

Translate. $lo g_{9} 3 = ?$ means " $9$ to what power is $3$ ?" Why this step? Reflex — convert to the question the log is asking.
Spot the root relationship. $3$ is the square root of $9$ , and a square root is the power $\frac{1}{2}$ : $9^{1/2} = 9 = 3$ . Why this step? Fractional exponents are roots ( $b^{1/ n} = n b$ ). When the target is a root of the base, the answer is a unit fraction.
Read off. $lo g_{9} 3 = \frac{1}{2}$ . Why this step? The value of the log is the exponent matched in step 2; since $9^{1/2} = 3$ , that exponent $\frac{1}{2}$ is the answer.

$lo g_{9} 3 = \frac{1}{2}$

Verify: $9^{1/2} = 9 = 3$ . ✓ Units check: exponents are pure numbers, and $\frac{1}{2}$ is between $0$ and $1$ as forecast (target between $1$ and base).

Worked example Example 5 — Case E: a base between 0 and 1 (decreasing curve)

Evaluate $lo g_{1/2} 8$ .

Forecast: here the base $\frac{1}{2}$ is less than $1$ (but still legal: $\frac{1}{2} > 0$ and $\frac{1}{2} \neq = 1$ ). Raising a number below $1$ to a positive power makes it smaller, so to grow up to $8$ we will need a negative exponent. This flips the intuition of Examples 1–3.

Translate. $lo g_{1/2} 8 = ?$ means " $\frac{1}{2}$ to what power is $8$ ?" Why this step? Reflex.
Rewrite the base as a power of $2$ . $\frac{1}{2} = 2^{- 1}$ , so $(\frac{1}{2})^{x} = 2^{- x}$ . Why this step? Turning the awkward base into a familiar one ( $2$ ) lets us reuse the powers-of- $2$ we already know.
Set the exponentials equal. We want $2^{- x} = 8 = 2^{3}$ , so $- x = 3$ , giving $x = - 3$ . Why this step? Once both sides share the same base, equal values force equal exponents — this is the answer because the log's value is exactly the $x$ that solves $(\frac{1}{2})^{x} = 8$ .

$lo g_{1/2} 8 = - 3$

Verify: $(\frac{1}{2})^{- 3} = 2^{3} = 8$ . ✓ Look at the figure: the $y = (1/2)^{x}$ curve goes downhill, so the height $8$ (above $1$ ) sits to the left at $x = - 3$ — the mirror image of the increasing case.

Worked example Example 6 — Case F: degenerate and edge INPUTS

Explain the status of (a) $lo g_{2} 0$ , (b) $lo g_{2} (- 8)$ , and describe what $lo g_{2} y$ does at both ends: (c) as $y \to 0^{+}$ and (d) as $y \to \infty$ .

Forecast: the domain of $lo g_{b}$ is $y > 0$ . So (a) and (b) should have no answer at all; (c) should run off to $- \infty$ ; and (d) should climb to $+ \infty$ (but slowly). Confirm each with the exponential.

(a) Zero input. Ask " $2$ to what power is $0$ ?" But $2^{x}$ is always positive — it can get microscopically close to $0$ yet never equals $0$ . No exponent works, so $lo g_{2} 0$ is undefined. Why this step? $0$ is the horizontal asymptote of $2^{x}$ ; the curve approaches but never lands there.
(b) Negative input. Ask " $2$ to what power is $- 8$ ?" Again $2^{x} > 0$ for every real $x$ , so no power is ever negative. $lo g_{2} (- 8)$ is undefined. Why this step? The sign of an exponent controls size, never the sign of the output — a common trap (see Mistake A in the parent note).
(c) Lower edge $y \to 0^{+}$ . As $y$ shrinks toward $0$ (e.g. $\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \dots$ ), the exponent runs $- 1, - 2, - 3, \dots$ downward without bound: $lo g_{2} y \to - \infty$ . Why this step? This is exactly the vertical asymptote $x = 0$ of the log graph, mirroring the horizontal asymptote of the exponential.
(d) Upper edge $y \to \infty$ . As $y$ grows huge (e.g. $2, 4, 8, 16, \dots$ ), the exponent runs $1, 2, 3, 4, \dots$ upward without bound: $lo g_{2} y \to + \infty$ . There is no ceiling — but it climbs slowly, since $y$ must double to raise the log by just $1$ . Why this step? The exponential $2^{x}$ rises to $+ \infty$ , so its inverse can accept ever-larger inputs and keep returning larger answers — the right-hand end of the s02 log curve.

Verify: for (c), $lo g_{2} (2^{- 10}) = - 10$ and $2^{- 10} = \frac{1}{1024}$ — a tiny positive input, large negative output. ✓ For (d), $lo g_{2} (2^{20}) = 20$ while $2^{20} = 1, 048, 576$ — you needed a million-fold input to reach $20$ . ✓ On the s01 figure, the black ✕ labelled "Case F" sits on the dashed line $y = 0$ precisely because no point of the curve matches that height; the s02 figure shows both runaway ends.

Worked example Example 6b — Case G: degenerate BASE

Explain why (a) $lo g_{1} 5$ , (b) $lo g_{1} 1$ , and (c) $lo g_{- 2} 8$ are all illegal.

Forecast: the base must satisfy $b > 0$ and $b \neq = 1$ . Every part below breaks one of those, so none is a valid logarithm — but for different reasons.

(a) Base $= 1$ , target $\neq = 1$ . Ask " $1$ to what power is $5$ ?" Since $1^{x} = 1$ for every $x$ , no power ever reaches $5$ . No answer exists ⇒ undefined. Why this step? $1^{x}$ is a flat line at height $1$ ; it can never output $5$ , so the inverse question is unanswerable.
(b) Base $= 1$ , target $= 1$ . Ask " $1$ to what power is $1$ ?" Now every $x$ works: $1^{0} = 1, 1^{99} = 1, \dots$ . Infinitely many answers ⇒ still not a well-defined function. Why this step? An inverse must return one value; a question with infinitely many answers cannot define a function. This is the second reason $b \neq = 1$ is required.
(c) Base $= - 2$ (negative). Ask " $- 2$ to what power is $8$ ?" $(- 2)^{1} = - 2, (- 2)^{2} = 4, (- 2)^{3} = - 8$ — the output flails between signs, and $(- 2)^{1/2} = - 2$ isn't even real. There is no smooth one-to-one curve to invert, so $lo g_{- 2}$ is not defined. Why this step? A negative base makes $b^{x}$ jump signs and go complex, destroying the one-to-one property an inverse needs.

$lo g_{1} 5, lo g_{1} 1, lo g_{- 2} 8 are all undefined.$

Verify: $1^{x} = 1$ never equals $5$ (part a); $1^{0} = 1^{5} = 1$ shows non-uniqueness (part b); $(- 2)^{1/2}$ is not a real number (part c). All three fail the legality rules $b > 0, b \neq = 1$ . ✓

Worked example Example 7 — Case H: cancellation identities in disguise

Simplify (a) $5^{l o g_{5} 12}$ , (b) $lo g_{3} (3^{7})$ , and (c) $1 0^{2 l o g_{10} 3}$ .

Forecast: the two golden identities are $b^{l o g_{b} y} = y$ and $lo g_{b} (b^{x}) = x$ . The first two are direct; the third is dressed up to look scary.

Before part (c) we need one small tool, so here it is self-contained:

(a) $5^{l o g_{5} 12}$ : the exponent $lo g_{5} 12$ is defined as "the power of $5$ that gives $12$ ." Raising $5$ to it gives $12$ . Why this step? This is $b^{l o g_{b} y} = y$ read straight off.
(b) $lo g_{3} (3^{7})$ : asks " $3$ to what power is $3^{7}$ ?" The exponent is sitting right there: $7$ . Why this step? This is $lo g_{b} (b^{x}) = x$ .
(c) $1 0^{2 l o g_{10} 3}$ : use the power rule just derived to pull the $2$ inside: $2 lo g_{10} 3 = lo g_{10} (3^{2}) = lo g_{10} 9$ . Then $1 0^{l o g_{10} 9} = 9$ . Why this step? We reshaped the exponent so a cancellation identity applies. (This same power rule is developed further in Laws of logarithms — product, quotient, power rules.)

$5^{l o g_{5} 12} = 12, lo g_{3} (3^{7}) = 7, 1 0^{2 l o g_{10} 3} = 9$

Verify: (a) $5^{l o g_{5} 12} = 12$ ✓; (b) $lo g_{3} 3^{7} = 7$ ✓; (c) $3^{2} = 9$ and $1 0^{l o g_{10} 9} = 9$ ✓.

Worked example Example 8 — Case I: a real-world word problem

A colony of bacteria doubles every hour. It starts with $1$ cell. After how many whole hours does it first exceed $1000$ cells?

Forecast: doubling means base $2$ . We want the exponent $t$ with $2^{t} \geq 1000$ . Since $2^{10} = 1024$ , guess around $10$ hours.

Model the story. After $t$ hours the count is $N = 2^{t}$ (double each hour, starting from $1$ ). Why this step? "Doubles every hour" is precisely repeated multiplication by $2$ — an exponential with base $2$ .
Set up the log question. We need the exponent $t$ making $2^{t} = 1000$ , i.e. $t = lo g_{2} 1000$ . Why this step? We know the answer ( $1000$ ) and the base ( $2$ ) and want the exponent — that is exactly what a log delivers.
Bracket the value between known powers. $2^{9} = 512$ (too small) and $2^{10} = 1024$ (just over). So $lo g_{2} 1000$ lies between $9$ and $10$ . Why this step? Trapping the target between two clean powers pins the answer without a calculator — the log itself is not a whole number, but the count only jumps at whole hours.
Interpret for whole hours. At $t = 9$ hours the colony is only $512$ cells (below $1000$ ); at $t = 10$ hours it is $1024$ cells (above $1000$ ). So it first exceeds $1000$ at the $10$ th hour. Why this step? The question asks for whole hours, so we report the first integer $t$ whose count clears the threshold.

$Cells first exceed 1000 at t = 10 hours.$

Verify: $2^{9} = 512 < 1000$ and $2^{10} = 1024 > 1000$ . ✓ Units: $t$ is in hours; the exponent of a "per-hour" doubling is a number of hours, as required.

Worked example Example 9 — Case J: exam twist, solve for the base

Find the base $b$ if $lo g_{b} 81 = 4$ .

Forecast: here the unknown is the base, not the exponent. The statement says " $b$ to the power $4$ is $81$ ", so $b$ is the fourth root of $81$ . Guess $3$ .

Convert to exponential form. $lo g_{b} 81 = 4 ⟺ b^{4} = 81$ . Why this step? The definition $lo g_{b} y = x ⟺ b^{x} = y$ works no matter which slot is unknown — here we solve for $b$ .
Undo the power. Take the positive fourth root: $b = 8 1^{1/4} = 481$ . Why this step? A fourth power is undone by a fourth root, and the base must be positive ( $b > 0$ ), so we keep only the positive root.
Evaluate. $3^{4} = 81$ , so $481 = 3$ , giving $b = 3$ . Why this step? $481$ is just the number whose fourth power is $81$ ; testing $3^{4} = 81$ confirms $b = 3$ (and $3 > 0, 3 \neq = 1$ , so it is a legal base).

$b = 3$

Verify: check the original statement — $lo g_{3} 81 = 4$ because $3^{4} = 81$ . ✓ And $b = 3$ satisfies the legality rules $b > 0, b \neq = 1$ .

Recall Active recall — cover the answers

$lo g_{2} 16$ ? ::: $4$
$lo g_{7} 1$ ? ::: $0$
$lo g_{2} \frac{1}{8}$ ? ::: $- 3$
$lo g_{9} 3$ ? ::: $\frac{1}{2}$
$lo g_{1/2} 8$ ? ::: $- 3$
Why is $lo g_{2} 0$ undefined? ::: $2^{x}$ never equals $0$ ; it only approaches it (asymptote).
What does $lo g_{2} y$ do as $y \to \infty$ ? ::: It rises to $+ \infty$ , but slowly (input must double to add $1$ ).
Why is $lo g_{1} 5$ undefined? ::: $1^{x} = 1$ always, so no power gives $5$ (base $1$ is illegal).
Why is $lo g_{- 2} 8$ undefined? ::: A negative base flips sign / goes complex, so it is not one-to-one.
Simplify $1 0^{2 l o g_{10} 3}$ . ::: $9$
Solve $lo g_{b} 81 = 4$ . ::: $b = 3$

Connections

Logarithm — definition as inverse of exponential (parent)
Exponential functions — $b^x$ and its graph
Laws of logarithms — product, quotient, power rules
Change of base formula
Solving exponential equations
Inverse functions and reflection in $y=x$