Change of base formula — proof: Worked examples

Everything rests on one fact from the Definition of a logarithm:

The scenario matrix

Before any example, here is the full map of cases this topic can throw at you. Each later example is tagged with the cell it covers.

#	Case class	What's special about it	Example
A	Clean integer answer	$x$ is an exact power of $a$	Ex 1
B	Non-integer answer	$x$ is not a power of $a$ → decimal, need calculator	Ex 2
C	Answer between 0 and 1	$x > 1$ but base bigger than $x$	Ex 3
D	Negative answer, base $> 1$	argument $x < 1$ (a fraction), base above 1	Ex 4
E	Fractional / reciprocal base, $x > 1$	old base $a$ like $\frac{1}{2}$ , argument above 1	Ex 5
E $^{'}$	Both below 1: $0 < a < 1$ and $0 < x < 1$	fraction base and fraction argument → answer flips back positive	Ex 5b
F $_{1}$	Degenerate arguments: $x = 1$ (any base) / base $= x$	$lo g_{a} 1 = 0$ for both $a > 1$ and $0 < a < 1$ ; $lo g_{a} a = 1$	Ex 6
F $_{2}$	Forbidden base $a = 1$	division by zero → formula collapses, $lo g_{1} x$ undefined	Ex 6b
G	Combining two logs (product/quotient cancels)	change of base as a simplifier	Ex 7
H	Word problem (real-world exponential)	translate a story into $lo g$	Ex 8
I	Exam twist (unknown base as variable)	solve for a base using change of base + Laws of logarithms	Ex 9

Intuition The sign of

every answer, in two pictures Left picture — base $a > 1$ (rising curve): $y = lo g_{a} x$ crosses zero at $x = 1$ .

$x > 1$ → above zero → answer positive.
$x = 1$ → exactly zero (because $a^{0} = 1$ ).
$0 < x < 1$ → below zero → answer negative.

Right picture — base $0 < a < 1$ (falling curve): same crossing at $x = 1$ , but the curve now slopes downward, so the two regions swap sign:

$x > 1$ → answer negative.
$x = 1$ → still exactly zero (the crossing never moves — see Ex 6).
$0 < x < 1$ → answer positive.

Alt-text: Two side-by-side plots of

y = lo g_{a} x

against

x

. Left plot (base

a > 1

) is a rising lavender curve passing through the point

(1, 0)

; the region

x > 1

is shaded mint and labelled POSITIVE, the region

0 < x < 1

is shaded butter and labelled NEGATIVE. Right plot (base

0 < a < 1

) is a falling lavender curve through the same point

(1, 0)

; now

x > 1

is shaded butter/NEGATIVE and

0 < x < 1

is shaded mint/POSITIVE — the sign regions are mirror-flipped. A red dotted vertical line marks

x = 1

(the zero crossing) in both plots.

Figure 1 — Sign of $lo g_{a} x$ . Left: a base bigger than 1 gives a rising curve (positive right of $x = 1$ , negative left). Right: a base between 0 and 1 gives a falling curve, so the shaded positive/negative regions are mirror-flipped. The red dotted line $x = 1$ is the zero crossing in both — it never moves, which is why $lo g_{a} 1 = 0$ for every allowed base.

Keep both curves in your head: which side is positive depends on whether the base is above or below 1.

Worked examples

Worked example Example 5 — Cell E: fractional / reciprocal base, argument above 1

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

Forecast: The base $\frac{1}{2}$ is less than 1, so raising it to a positive power makes it smaller, not bigger. To reach a number $8 > 1$ we need a negative power. On the right (falling) picture, $x > 1$ sits in the negative region. Since $(\frac{1}{2})^{- 3} = 2^{3} = 8$ , guess $- 3$ .

Change of base to $b = 2$ (a smart choice — $\frac{1}{2}$ and $8$ are both powers of $2$ ): $lo g_{1/2} 8 = \frac{lo g _{2} 8}{lo g _{2} ( 1/2 )}$ . Why this step? Picking $b = 2$ makes both logs exact whole numbers — no decimals at all.
Top: $lo g_{2} 8 = 3$ . Bottom: $lo g_{2} (1/2) = - 1$ because $2^{- 1} = \frac{1}{2}$ . Why this step? Both are read straight off the definition $a^{y} = x$ .
Evaluate: $\frac{3}{- 1} = - 3$ .

Verify: $(\frac{1}{2})^{- 3} = 2^{3} = 8$ . ✓ A base below 1 with argument above 1 gives a negative answer — matches the right-hand picture.

Worked example Example 5b — Cell E′: BOTH below 1 (

$0 < a < 1$ and $0 < x < 1$ ) Compute $lo g_{1/2} \frac{1}{4}$ .

Forecast: Here the base $\frac{1}{2}$ is below 1 and the argument $\frac{1}{4}$ is below 1. On the right (falling) picture, the region $0 < x < 1$ is the positive region — so unlike Cell D, this answer comes out positive. Since $(\frac{1}{2})^{2} = \frac{1}{4}$ , guess $+ 2$ .

Change of base to $b = 2$ (both $\frac{1}{2}$ and $\frac{1}{4}$ are powers of $2$ ): $lo g_{1/2} \frac{1}{4} = \frac{lo g _{2} ( 1/4 )}{lo g _{2} ( 1/2 )}$ . Why this step? A common power-of-2 base makes both logs exact, exposing the sign cleanly.
Top: $lo g_{2} (1/4) = - 2$ because $2^{- 2} = \frac{1}{4}$ . Bottom: $lo g_{2} (1/2) = - 1$ . Why this step? Both arguments are below 1, so both logs are negative.
Evaluate: $\frac{- 2}{- 1} = + 2$ . Why this step? A negative over a negative is positive — this is the arithmetic reason the falling-curve region below $x = 1$ is positive.

Verify: $(\frac{1}{2})^{2} = \frac{1}{4}$ . ✓ Two fractions combine to a positive answer — the last unillustrated sign case.

Worked example Example 6 — Cell F₁: degenerate arguments (

$x = 1$ for BOTH kinds of base; base $= x$ ) Evaluate (a) $lo g_{7} 1$ , (b) $lo g_{1/3} 1$ , and (c) $lo g_{7} 7$ .

Forecast: (a) "what power of $7$ gives $1$ ?" — any base to the power $0$ is $1$ , so guess 0. (b) same question for a fractional base $\frac{1}{3}$ — the crossing point $x = 1$ never moves, so guess 0 again. (c) "what power of $7$ gives $7$ ?" — that's just 1.

(a) base above 1. Change of base: $lo g_{7} 1 = \frac{lo g _{10} 1}{lo g _{10} 7} = \frac{0}{0.8451} = 0$ . Why this step? The numerator $lo g_{10} 1 = 0$ because $1 0^{0} = 1$ ; a zero on top forces the whole answer to zero. This is the $x = 1$ crossing on the left (rising) picture.
(b) base below 1. Change of base: $lo g_{1/3} 1 = \frac{lo g _{10} 1}{lo g _{10} ( 1/3 )} = \frac{0}{- 0.4771} = 0$ . Why this step? The top is still $lo g_{10} 1 = 0$ ; the denominator being negative (base $< 1$ ) doesn't matter — $0$ divided by anything nonzero is $0$ . This is the same crossing point on the right (falling) picture, confirming it never shifts. So the previously un-drilled quadrant (base $< 1$ , $x = 1$ ) also gives 0.
(c) base $=$ argument. Change of base: $lo g_{7} 7 = \frac{lo g _{10} 7}{lo g _{10} 7} = 1$ . Why this step? Identical top and bottom → the ratio is $1$ . Whenever argument equals base, the answer is $1$ .

Verify: $7^{0} = 1$ ✓, $(\frac{1}{3})^{0} = 1$ ✓, and $7^{1} = 7$ ✓. The zero crossing at $x = 1$ holds for every allowed base.

Worked example Example 6b — Cell F₂: the forbidden base

$a = 1$ Attempt $lo g_{1} 5$ . Why is it undefined?

Forecast: "what power of $1$ gives $5$ ?" — $1$ raised to any power is still $1$ , so no power ever reaches $5$ . Expect "no answer / undefined".

Try change of base to $b = 10$ : $lo g_{1} 5 = \frac{lo g _{10} 5}{lo g _{10} 1} = \frac{0.6990}{0}$ . Why this step? Feeding $a = 1$ into the formula puts $lo g_{10} 1 = 0$ in the denominator — a division by zero, which is undefined. This is the algebraic image of "no power works".
Confirm from the definition: $1^{y} = 5$ has no solution, since $1^{y} = 1$ for every real $y$ . Why this step? Independently of any calculator, the question itself has no answer — the base $1$ is flat, it never grows or shrinks.

Verify: For any target $x \neq = 1$ , $lo g_{1} x$ is undefined; and $lo g_{1} 1$ is ambiguous ( $1^{y} = 1$ for all $y$ ). This is exactly why the domain rule bans $a = 1$ , and why the parent proof insisted on $a \neq = 1$ before dividing at Step 5. ✗ (correctly rejected)

Worked example Example 8 — Cell H: real-world word problem

A colony of bacteria triples every hour, starting from $500$ . After how many hours does it first exceed $500000$ ? (Links to Solving exponential equations.)

Forecast: Multiplying by $3$ repeatedly, $500 \to 1500 \to 4500 \to \dots$ ; a thousand-fold rise needs roughly $lo g_{3} 1000 \approx 6$ – $7$ hours. Guess about 7.

Model: after $t$ hours the count is $500 \cdot 3^{t}$ . Set $500 \cdot 3^{t} = 500000$ . Why this step? Tripling each hour is repeated multiplication = an exponential with base $3$ .
Divide by $500$ : $3^{t} = 1000$ . In log form: $t = lo g_{3} 1000$ . Why this step? The Definition of a logarithm turns "unknown exponent" into a log.
Change of base to compute it: $t = \frac{lo g _{10} 1000}{lo g _{10} 3} = \frac{3}{0.4771} \approx 6.288$ . Why this step? No base-3 button, so translate to base 10; and $lo g_{10} 1000 = 3$ exactly because $1 0^{3} = 1000$ .
Since $t$ must be a whole hour to have passed, it first exceeds at $t = 7$ hours. Why this step? At $t = 6.288$ it just crosses; the next completed hour is $7$ .

Verify: $500 \cdot 3^{6} = 500 \cdot 729 = 364500 < 500000$ , but $500 \cdot 3^{7} = 500 \cdot 2187 = 1, 093, 500 > 500000$ . ✓ First exceeds at hour 7.

Worked example Example 9 — Cell I: exam twist (base is the unknown)

Solve $lo g_{a} 16 = 2$ for the base $a$ (with $a > 0, a \neq = 1$ ).

Forecast: " $a$ squared gives $16$ " points straight at $a = 4$ .

Rewrite using the definition: $lo g_{a} 16 = 2 ⟺ a^{2} = 16$ . Why this step? When the base is unknown, going to exponential form is cleaner than change of base — but let's also confirm with change of base to show consistency.
Cross-check via change of base: $lo g_{a} 16 = \frac{lo g _{10} 16}{lo g _{10} a} = 2 ⟹ lo g_{10} a = \frac{1}{2} lo g_{10} 16 = lo g_{10} 4$ . Why this step? Isolating $lo g_{10} a$ and using $\frac{1}{2} lo g_{10} 16 = lo g_{10} 16 = lo g_{10} 4$ (a power law) pins down $a$ .
Therefore $a = 4$ . (Reject $a = - 4$ : the domain rule says bases must be positive.)

Verify: $4^{2} = 16$ , so $lo g_{4} 16 = 2$ . ✓ We correctly discard the negative root because a logarithm base must be positive and $\neq = 1$ .

Recall Which cell was hardest — sign-check drill

For each, state the sign of the answer before computing: $lo g_{5} 0.2$ ::: negative (base $> 1$ , argument $< 1$ ) $lo g_{0.5} 4$ ::: negative (base $< 1$ , argument $> 1$ ) $lo g_{0.5} 0.25$ ::: positive (base $< 1$ AND argument $< 1$ → signs flip back) $lo g_{10} 1$ ::: exactly zero (argument $= 1$ ) $lo g_{1/3} 1$ ::: exactly zero (argument $= 1$ , even for a base below 1) $lo g_{1} 5$ ::: undefined (forbidden base $a = 1$ ) $lo g_{3} 9$ ::: positive (base $> 1$ , argument $> 1$ )

Active recall

State the domain rules for

lo g_{a} x

a > 0

a \neq = 1

, and

x > 0

Which cell gives a negative answer from a base below 1?

Cell E — e.g.

lo g_{1/2} 8 = - 3

(argument above 1).

When base and argument are BOTH below 1, what is the sign?

Positive — e.g.

lo g_{1/2} (1/4) = + 2

Value of

lo g_{7} 1

0

, since

7^{0} = 1

Value of

lo g_{1/3} 1

0

— the crossing at

x = 1

holds for a base below 1 too.

Why is

lo g_{1} 5

undefined?

The formula divides by

lo g_{b} 1 = 0

; and

1^{y} = 5

has no solution.

Value of

lo g_{7} 7

1

, since

7^{1} = 7

lo g_{9} 3 = ?

0.5

(because

9^{1/2} = 3

lo g_{2} (1/8) = ?

- 3

lo g_{1/2} (1/4) = ?

+ 2

Solve

lo g_{a} 16 = 2

a = 4

(reject

- 4

lo g_{8} x \cdot lo g_{x} 64 = ?

2

First hour bacteria exceed

500000

(triple hourly from

500

hour

7

Connections

Change of base formula — proof — the parent proof these examples exercise.
Definition of a logarithm — used in Cells F₁, F₂ and I to switch to exponential form, and to justify the domain rules.
Laws of logarithms — power law drives Cells D, E′, G, I.
Exponential functions — the model behind the word problem and the falling-curve picture.
Natural logarithm ln — the base- $e$ ruler in Cell B.
Solving exponential equations — the workflow of Cell H.