Change of base formula — proof: Visual walkthrough

Before any symbol, one promise: a logarithm is a question. When we write $lo g_{a} x$ we are asking "what power must I raise the base $a$ to, in order to get the number $x$ ?" The answer is a power (an exponent), and that power need not be a whole number.

Step 0 — The single seed fact (a picture of what a log is)

WHAT. The whole proof grows from one fact, valid for $a > 0, a \neq = 1, x > 0$ : $the question lo g_{a} x = the answer y ⟺ raise base a to power y a^{y} = the target number x .$

Term by term: $a$ is the base, $y$ is the exponent (any real number, possibly negative), $x$ is the result. The two-headed arrow $⟺$ means these are the same sentence written two ways.

WHY. We cannot do algebra on a question. But we can do algebra on the exponential form $a^{y} = x$ . This flip is our only tool for turning a log into something we can push around.

PICTURE. The left panel shows the log as a "what power?" question; the right panel shows the smooth ramp $a^{y}$ with whole-number landmarks dotted on it — the ramp fills in every value in between.

See Definition of a logarithm — this seed fact is the one thing you must trust.

Step 1 — Name the mystery

WHAT. Let $y = lo g_{a} x .$ Here $y$ is just a nickname for the number we are hunting — a signed position on the ramp.

WHY. A mystery you can name is a mystery you can chase. Calling the answer $y$ lets us write equations about it instead of staring at the log symbol. Nothing has been assumed — $y$ is defined to be $lo g_{a} x$ .

PICTURE. We mark the unknown signed height $y$ on the smooth ramp from Step 0, including the region where $y$ dips below zero.

Step 2 — Flip to exponential form

WHAT. Apply the seed fact to $y = lo g_{a} x$ : $a^{y} = x .$

WHY. This is the flip promised in Step 0. Logs are hard to manipulate directly; exponentials are not, because a log of any base can be dropped onto a power and it will slide the exponent down (that's the power law, coming in Step 4). We convert now so that later we have an exponent to grab.

PICTURE. The exponent $y$ is the horizontal position on the smooth ramp $a^{y}$ whose height reaches $x$ . When $x < 1$ that position sits to the left of zero ( $y < 0$ ). This is Exponential functions read backwards.

Step 3 — Measure both sides with a NEW ruler ( $lo g_{b}$ )

WHAT. Take $lo g_{b}$ of both sides: $lo g_{b} (a^{y}) = lo g_{b} x .$

Term by term: $lo g_{b} (\cdot)$ is our new ruler. For this new ruler to be a valid logarithm at all, its base must obey the very same domain rules from the top of the page: $b > 0$ and $b \neq = 1$ . Any such $b$ works, but in practice we pick a base our calculator understands — usually $10$ or the natural log base $e$ . We apply $lo g_{b}$ to the left side $a^{y}$ and the right side $x$ ; both are positive (Step 2), so both are legal arguments.

WHY. Equality is a balance scale: if you do the same thing to both pans, it stays level. We choose the operation $lo g_{b}$ deliberately — it is exactly the tool that will let us peel the exponent $y$ off in the next step. Any other function (squaring, adding) would keep $y$ trapped in the exponent.

PICTURE. Both sides of the equation get "photographed" through the same base- $b$ measuring lens.

Step 4 — The power law drags the exponent down

WHAT. Use the power law $lo g_{b} (a^{y}) = y lo g_{b} a$ : $y lo g_{b} a = lo g_{b} x .$

Term by term: the exponent $y$ has jumped down in front to become a multiplier; $lo g_{b} a$ is a plain fixed number (the signed height of the old base $a$ on the new ruler); $lo g_{b} x$ is another plain number (the signed height of $x$ on the new ruler).

WHY. This is the heartbeat of the whole proof. A logarithm's superpower is turning powers into products — see Laws of logarithms. That is why we bothered flipping to exponential form in Step 2: so that Step 4 could liberate $y$ from the roof of the power and set it down as an ordinary factor we can divide away.

PICTURE. The little exponent $y$ literally slides down and becomes a coefficient.

Step 5 — Isolate $y$ (and check the division is legal)

WHAT. Divide both sides by $lo g_{b} a$ : $y = \frac{l o g _{b} x}{l o g _{b} a} .$

WHY — and the degenerate case. You may only divide by something that is not zero. When is $lo g_{b} a = 0$ ? Exactly when $a = 1$ , because $b^{0} = 1$ forces the argument to be $1$ . But a logarithm's base is never allowed to be $1$ (a base of $1$ can never climb — $1^{anything} = 1$ ), so $a \neq = 1$ is baked into the rules. Hence $lo g_{b} a \neq = 0$ and the division is always safe. No forbidden move ever happens.

PICTURE. The scale is rebalanced by cutting both pans by the same factor $lo g_{b} a$ ; the panel also shows the "danger zone" $a = 1$ crossed out.

Step 6 — Remember what $y$ was

WHAT. In Step 1 we set $y = lo g_{a} x$ . Substitute it back: $lo g_{a} x = \frac{lo g _{b} x}{lo g _{b} a} ■$

Term by term: argument $x$ on top, old base $a$ on the bottom, and the new base $b$ on both logs. Valid for $a > 0, a \neq = 1$ ; $x > 0$ ; and any new base $b > 0, b \neq = 1$ .

WHY. The nickname has done its job; we restore the real name so the result is about the thing we actually wanted.

PICTURE. The final formula with each slot colour-coded to the role it plays.

Step 7 — Why any new base gives the same answer

WHAT. Pick $b = 10$ or $b = e$ or $b = 7$ (each a valid base, $> 0$ and $\neq = 1$ ) — the value of $\frac{l o g _{b} x}{l o g _{b} a}$ does not change. Concretely for $lo g_{2} 8$ : $\frac{l o g _{10} 8}{l o g _{10} 2} = 3, \frac{l n 8}{l n 2} = 3, \frac{l o g _{7} 8}{l o g _{7} 2} = 3.$

WHY. Changing $b$ multiplies both top and bottom by the same stretch factor, and a shared factor cancels. This is the "same power, different ruler" idea: rescaling the ruler cannot change how many 2's fit into 8. This is also what makes solving exponential equations possible on a real calculator.

PICTURE. Three rulers of different spacing, same measured ratio.

The one-picture summary

This single figure compresses all seven steps into one flow: seed fact → name → flip → measure → power law → divide → restore, with the "same ruler cancels" idea riding alongside.

Recall Feynman: tell the whole walkthrough to a friend

"I want $lo g_{2} 8$ — what power turns 2 into 8 — but my calculator only speaks base 10. So I give the unknown answer a name, $y$ . Saying ' $y$ is that log' is the same as saying $2^{y} = 8$ . Now I photograph both sides through my base-10 camera: $lo g_{10} (2^{y}) = lo g_{10} 8$ . The magic law lets the exponent $y$ slide down front: $y lo g_{10} 2 = lo g_{10} 8$ . I divide by $lo g_{10} 2$ — safe, because base 2 isn't 1 — and get $y = \frac{l o g _{10} 8}{l o g _{10} 2}$ . Put the name back: $lo g_{2} 8 = \frac{l o g _{10} 8}{l o g _{10} 2} = 3$ . Two reminders as I go: the power $y$ needn't be a whole number (for $lo g_{5} 40$ it's about $2.29$ , landing between the ramp's whole steps), and it can be negative (for $lo g_{2} \frac{1}{4}$ it's $- 2$ ). And if my friend used base $e$ instead, they'd get the identical answer, because switching rulers stretches top and bottom the same amount and it cancels."

Solutions

(1)

lo g_{a} x = y \Leftrightarrow a^{y} = x

for

a > 0, a \neq = 1, x > 0

. (2) So we can apply a log and use the power law. (3) The power law

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

. (4) Because

a \neq = 1

, so

lo g_{b} a \neq = 0

. (5)

b > 0

and

b \neq = 1

, so that

lo g_{b}

is itself a valid logarithm. (6) e.g.

x = \frac{1}{4}

lo g_{2} \frac{1}{4} = - 2

since

2^{- 2} = \frac{1}{4}

(values in

(0, 1)

need a negative exponent).

Connections

Change of base formula — proof — the parent this walkthrough illustrates.
Definition of a logarithm — the Step 0 seed fact and the domain rules.
Exponential functions — the Step 2 flip.
Laws of logarithms — the Step 4 power law.
Natural logarithm ln — the usual new base $b = e$ .
Solving exponential equations — where this translator earns its keep.

Visual walkthrough — Change of base formula — proof