Change of base formula — proof: Question bank

Reminder of the formula we are probing (from the parent Change of Base — proof): $lo g_{a} x = \frac{l o g _{b} x}{l o g _{b} a}, a, b > 0, a \neq = 1, b \neq = 1, x > 0.$

The derivation, inline (so this page stands alone)

You should not have to leave to justify the traps. Here is the full proof in six lines; every "Why question" below points back to one of these steps.

The visual anchor — two rulers, same height

The whole formula is a ruler-swap. The picture below is the anchor for the metaphor used all over this page: measuring the same "power-height" with a base-10 ruler and a base- $e$ ruler, then dividing to cancel the ruler.

Left panel (green bar): the value $lo g_{2} 8 = 3$ shown as the true "height" of the question "how many 2s make 8?".
Middle panel: two different rulers measuring the same pair of numbers. The blue bars are the base-10 ruler ( $lo g_{10} 8$ and $lo g_{10} 2$ ); the yellow bars are the base- $e$ ruler ( $ln 8$ and $ln 2$ ). Notice the blue tick heights differ from the yellow ones — different rulers give different tick heights for the same numbers.
Right panel: dividing each ruler's two heights. The blue bar is the base-10 ratio $lo g_{10} 8/ lo g_{10} 2$ (this is the numerator over denominator); the yellow bar is the base- $e$ ratio $ln 8/ ln 2$ . The ruler's scale factor cancels, so both bars land on the red dashed line at $3$ — the choice of new base never appears in the answer.

Second anchor — read the axes first. The horizontal axis is the new base $b$ , swept continuously from $b = 1.2$ up to $b = 12$ (every value along the axis is a different choice of new base). The vertical axis is the value of each log.

The blue curve is the numerator $lo g_{b} 8$ ; the yellow curve is the denominator $lo g_{b} 2$ . Both shrink as $b$ grows — a bigger ruler makes every tick shorter.
The red curve is their ratio $lo g_{b} 8/ lo g_{b} 2$ , and it is a flat horizontal line at $3$ : no matter which new base $b$ you slide to along the axis, the ratio never moves. Any "Why questions" trap about the choice of $b$ is answered by this flat red line.

True or false — justify

lo g_{2} 8

and

\frac{lo g _{10} 8}{lo g _{10} 2}

are different numbers because the second uses base 10.

False — both equal

3

. Change of base doesn't change the answer to "what power gives 8?"; it only changes the ruler you measure the two heights with, and the scaling cancels (right panel of figure s01: both coloured bars hit

3

The new base

b

must be

10

e

for the formula to be valid.

False — the proof used a completely general

b

; any

b > 0, b \neq = 1

works. We prefer

10

e

only because calculators have those buttons.

You may choose the new base

b

to equal the old base

a

False — if

b = a

the formula gives

\frac{l o g _{a} x}{l o g _{a} a} = \frac{l o g _{a} x}{1}

, which is a true but useless identity; more importantly you'd never do this to escape base

a

. It's allowed algebraically but pointless.

lo g_{a} b \cdot lo g_{b} a = 1

for every valid

a, b

True — by the reciprocal identity

lo g_{a} b = 1/ lo g_{b} a

, so their product is

1

(provided

a, b > 0

and neither is

1

). They are two measurements of the same relationship, one flipped.

x = 1

then

lo g_{a} x = 0

no matter what base you convert to.

True —

lo g_{b} 1 = 0

in every base since

b^{0} = 1

, so the numerator

lo g_{b} x

0

and the whole fraction is

0

. The trap is thinking the base affects a value that is genuinely

0

Choosing a larger new base

b

makes

lo g_{a} x

come out larger.

False — the value

lo g_{a} x

is fixed; only the numerator and denominator both grow together and their ratio stays constant (the flat red line in figure s02, as

b

sweeps along the horizontal axis). The output is independent of

b

by design.

\frac{lo g _{b} x}{lo g _{b} a}

is undefined whenever

lo g_{b} a

could be negative.

False — a negative denominator is perfectly fine; the fraction is only undefined when

lo g_{b} a = 0

, i.e. when

a = 1

, which is already banned as a base.

Spot the error

lo g_{a} x = \frac{lo g a}{lo g x}

Two errors: the base

a

is written on top (it belongs on the bottom), and each

lo g

lost its argument. Correct (choosing new base

b

\frac{lo g _{b} x}{lo g _{b} a}

, with argument

x

on top and old base

a

on the bottom — both logs carry the same new base

b

lo g_{2} 8 = \frac{lo g _{10} 2}{lo g _{10} 8}

Numerator and denominator are swapped. This gives

\approx 0.33

, but

lo g_{2} 8 = 3

. The correct base-10 version is

\frac{lo g _{10} 8}{lo g _{10} 2}

. Sanity check:

8

is bigger than

2

, so its log must be on top to give an answer

> 1

lo g_{a} x = \frac{lo g _{a}}{lo g x}

— just split the single log."

A single log

lo g_{a} x

does not split;

lo g_{a}

alone is meaningless because a log needs both a base and an argument. The correct object is

\frac{lo g _{b} x}{lo g _{b} a}

: two complete logs, each in the same new base

b

"Take

lo g_{b}

of both sides of

a^{y} = x

to get

lo g_{b} a \cdot lo g_{b} y = lo g_{b} x

The power law puts the exponent out front as a multiplier of the base's log:

lo g_{b} (a^{y}) = y lo g_{b} a

, not

lo g_{b} a \cdot lo g_{b} y

. You never take a second log of

y

. (This is Step 4 of the inline proof.)

lo g_{a} x = lo g_{b} x - lo g_{b} a

because logs turn division into subtraction."

The subtraction law

lo g_{b} (x / a) = lo g_{b} x - lo g_{b} a

is a different identity. Change of base needs division of two logs, i.e.

\frac{lo g _{b} x}{lo g _{b} a}

, not the log of a quotient — the power law produced a coefficient

y lo g_{b} a

, which we divide, not subtract.

"Since

ln 40/ ln 5 \approx 2.29

is not a whole number, the answer is only approximate and change of base failed."

The exact value

lo g_{5} 40

genuinely is irrational;

2.29

is a rounded decimal of the true value. Change of base is an exact identity — the approximation lives only in the decimal digits of

ln 40

and

ln 5

Why questions

Why do we begin Step 1 by naming the unknown

y = lo g_{a} x

instead of manipulating

lo g_{a} x

directly?

You can't do algebra on a phrase you can't rearrange; naming it

y

turns the question "what power?" into a single symbol we can move across an equals sign and eventually isolate. Every later step exists only to solve for that

y

Why must we take the log of both sides in Step 3 of the proof, not just one?

Applying the same function to both sides of

a^{y} = x

preserves the equality; touching only one side would create a false statement. Equality is a two-sided promise.

Why do we take a logarithm at Step 3 rather than, say, a square root?

The unknown

y

sits in the exponent of

a^{y}

. Only the log's power law can drag an exponent down into a multiplier; a root or any other operation leaves

y

stuck upstairs.

Why is dividing by

lo g_{b} a

guaranteed to be legal?

Because

a \neq = 1

forces

lo g_{b} a \neq = 0

(only

lo g_{b} 1 = 0

). Division by a nonzero quantity is always safe, so isolating

y

never risks divide-by-zero.

Why does the argument

x

end up on top and the base

a

on the bottom, not the reverse?

The exponent

y = lo g_{a} x

multiplied

lo g_{b} a

(Step 4:

y lo g_{b} a = lo g_{b} x

). Solving for

y

divides by that coefficient, sending

lo g_{b} a

to the denominator and leaving

lo g_{b} x

on top.

Why does the choice of new base

b

not appear in the final answer?

Both numerator and denominator scale by exactly the same base-conversion factor, so the ratio is invariant — it's the same power measured with two proportional rulers (the flat red ratio line in figure s02 as

b

sweeps the horizontal axis). This is why

10

and

e

give identical results.

Why does the reciprocal identity

lo g_{a} b = 1/ lo g_{b} a

pop out so easily?

Set

x = b

in the formula with new base

b

: the numerator becomes

lo g_{b} b = 1

(since

b^{1} = b

), leaving

1/ lo g_{b} a

. Swapping base and argument just flips the fraction.

Edge cases

What happens to the formula if you try

x = 1

lo g_{b} 1 = 0

lo g_{a} 1 = 0/ lo g_{b} a = 0

— correct, since

a^{0} = 1

for any base. No division problem: the zero is in the numerator, not the denominator.

What if the argument

x < 0

x = 0

— can change of base rescue it?

No.

lo g_{b} x

requires

x > 0

; there is no real power of a positive base giving a non-positive number. Change of base only translates an already-valid log, it can't extend the domain.

lo g_{1} x

ever meaningful, so that we might convert out of base

1

No — base

1

is banned because

1^{y} = 1

for all

y

, so "

lo g_{1} x

" would have to answer "what power turns

1

into

x

?", which has no answer (or infinitely many if

x = 1

). The formula's ban

a \neq = 1

enforces exactly this.

What if the old base is a proper fraction,

0 < a < 1

(say

lo g_{1/2} 8

) — is the formula still valid?

Yes —

a = 1/2

satisfies

a > 0, a \neq = 1

, so it is a legal old base. Here

lo g_{1/2} 8 = \frac{l o g _{b} 8}{l o g _{b} ( 1/2 )}

; the denominator

lo g_{b} (1/2)

is negative, so the result is negative (

= - 3

, since

(1/2)^{- 3} = 8

). The sign is handled automatically — no special rule needed.

For

0 < a < 1

, does

lo g_{a} x

increase or decrease as

x

grows?

It decreases — a fractional base means bigger arguments need more negative powers (e.g.

(1/2)^{- 1} = 2, (1/2)^{- 2} = 4

). Change of base reflects this correctly because

lo g_{b} a < 0

flips the sign of the whole fraction; the formula never assumes

a > 1

Can the new base

b

be a fraction like

1/2

Yes —

b = 1/2

satisfies

b > 0, b \neq = 1

, so it is valid. Both

lo g_{1/2} x

and

lo g_{1/2} a

come out negative-scaled, but their ratio still equals the true

lo g_{a} x

; the ruler being "upside down" cancels.

What is

lo g_{a} a

for any valid base, and why is it the smallest instructive edge case?

lo g_{a} a = 1

because

a^{1} = a

. It's the sanity check that any conversion must respect:

\frac{l o g _{b} a}{l o g _{b} a} = 1

, confirming the machinery doesn't corrupt trivial truths.

If both

a > 1

and

x < 1

(say

lo g_{2} 0.5

), does the formula still behave?

Yes —

lo g_{b} x

is negative (since

x < 1

) while

lo g_{b} a

is positive, giving a negative result.

lo g_{2} 0.5 = - 1

because

2^{- 1} = 0.5

; the signs handle themselves, no special rule needed.

Connections

Definition of a logarithm — the single fact every trap here tests.
Laws of logarithms — the power law and the quotient law that students confuse.
Exponential functions — why $x > 0$ is unavoidable in the domain edge cases.
Natural logarithm ln — the usual "new base" and why it isn't special.
Solving exponential equations — where these traps bite in practice.