Logarithm — definition as inverse of exponential

WHAT is a logarithm?

WHY the restrictions?

$b>0$ : negative bases like $(-2)^x$ jump between signs / go complex — not a smooth function.
$b\neq 1$ : because $1^x = 1$ always, so " $1$ to what power gives $5$ ?" has no answer — not invertible.
$y>0$ : since $b^x>0$ for every real $x$ , its inverse can only accept positive inputs.

Deriving it from scratch (no memorising)

Step 1. Start with the exponential $y=b^x$ . Why this step? We want the inverse, so we ask which $x$ produced a given $y$ .

Step 2. Swap roles: solve for $x$ in terms of $y$ . There is no algebra trick to isolate $x$ from an exponent using $+,-,\times,\div$ — so we invent a symbol for the answer and call it $\log_b y$ . Why this step? The exponent is "trapped upstairs"; the log is the tool built specifically to bring it down.

Step 3. By construction, $x=\log_b y$ . Substituting back into $y=b^x$ gives $y=b^{\log_b y}$ — our first golden identity. Why this step? Substitution proves the definition is self-consistent.

Step 4. Starting instead from $x$ : apply $b^{(\cdot)}$ then $\log_b$ : $\log_b(b^x)=x$ because the exponent that produces $b^x$ is literally $x$ . Why this step? Confirms cancellation works in both directions ⇒ genuine inverse.

Figure — Logarithm — definition as inverse of exponential

Worked examples

Common mistake Steel-manned common errors

Mistake A: " $\log$ of a negative number is fine, e.g. $\log_2(-8)=-3$ ." Why it feels right: students see the minus in $-3$ and think it "makes" the negative. Fix: $2^{-3}=\tfrac18>0$ . No real power of a positive base is ever negative, so $\log_2(-8)$ is undefined. The exponent's sign controls size, never the sign of the output.

Mistake B: " $\log_b(x+y)=\log_b x+\log_b y$ ." Why it feels right: it mimics distributive/linear rules we love. Fix: logs turn multiplication into addition, not addition into addition. $\log_b(xy)=\log_b x+\log_b y$ — but $\log_b(x+y)$ has no simplification. Test: $\log_{10}(1+1)=\log_{10}2\approx0.30$ , while $\log_{10}1+\log_{10}1=0$ . Not equal.

Mistake C: confusing $\log_b(b^x)=x$ with $(\log_b b)^x$ . Fix: $\log_b(b^x)$ is "the exponent giving $b^x$ " $=x$ . $\log_b b=1$ , so $(\log_b b)^x=1^x=1$ . Totally different.

Recall Feynman: explain to a 12-year-old

Imagine a magic doubling machine. Put in a $1$ , press the button $3$ times, and it grows $1\to2\to4\to8$ . The exponent is how many times you pressed the button. Now suppose your friend hands you an $8$ and asks, "How many times did you press?" Answering that is exactly what a logarithm does: it counts the button-presses. "Log base 2 of 8 = 3" just means "you pressed the doubling button 3 times." Because you can never shrink to zero or below by doubling from $1$ , you can only ask this about positive numbers.

Connections

Exponential functions — $b^x$ and its graph
Laws of logarithms — product, quotient, power rules
Change of base formula
Natural logarithm $\ln$ and Euler's number $e$
Solving exponential equations
Inverse functions and reflection in $y=x$

What does

\log_b y

literally represent?

The exponent to which

b

must be raised to obtain

y

(i.e. the solution

x

b^x=y

State the definition linking log and exponential form.

\log_b y = x \iff b^x = y

, for

b>0,\ b\neq1,\ y>0

Why must the base satisfy

b\neq 1

Because

1^x=1

for all

x

, so

b^x

is not one-to-one and has no inverse.

Why is the domain of

\log_b

only

y>0

Since

b^x>0

for every real

x

, the inverse can only accept positive inputs.

Simplify

b^{\log_b y}

y

(the two operations cancel; this holds for

y>0

Simplify

\log_b(b^x)

x

(for all real

x

Evaluate

\log_2 8

3

, because

2^3=8

Evaluate

\log_3 \tfrac{1}{9}

-2

, because

3^{-2}=\tfrac19

Reflecting

y=b^x

in which line gives

y=\log_b x

The line

y=x

\log_b(x+y)=\log_b x+\log_b y

true?

No. Logs convert products to sums:

\log_b(xy)=\log_b x+\log_b y

; sums have no such rule.

What point does every

\log_b x

graph pass through, and why?

(1,0)

, because

b^0=1

\log_b 1=0

What is the vertical asymptote of

y=\log_b x

x=0

(image of the horizontal asymptote

y=0

b^x

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, exponential ka kaam simple hai: base aur power do, answer nikaalo — jaise $2^3 = 8$ . Lekin kaafi baar ulta problem aata hai: humein answer ( $8$ ) aur base ( $2$ ) pata hai, magar power (kitni baar) nahi pata. Yahi "kitni power?" wala sawaal poochhne ka tool hai logarithm. Isliye kehte hain: $\log_2 8 = 3$ ka matlab hai " $2$ ko kis power tak le jaayein ki $8$ mile? — 3 power tak". Bas itni si baat: log ka matlab hai chhupa hua exponent.

Definition ek line mein: $\log_b y = x$ tabhi jab $b^x = y$ . Yeh dono ek hi baat ke do roop hain — ek exponential form, ek log form. Jab equation solve karni ho, in dono forms ke beech switch karna sabse powerful trick hai. Aur inverse hone ki wajah se do golden identities free milti hain: $b^{\log_b y} = y$ aur $\log_b(b^x) = x$ — ye ek doosre ko cancel kar dete hain, jaise plus aur minus.

Do restrictions yaad rakho, warna trap mein phasoge. Pehla: $y$ hamesha positive hona chahiye, kyunki $b^x$ kabhi zero ya negative nahi hota, isliye $\log$ of negative undefined hai. Doosra: base $b \neq 1$ , kyunki $1$ ki koi bhi power $1$ hi rehti hai, toh woh invertible nahi. Graph ke liye ek picture dimaag mein rakho: $y = \log_b x$ actually $y = b^x$ ka mirror image hai $y = x$ line ke around — bas $x$ aur $y$ axis swap ho gaye. Isse tumhe curve ka shape, asymptote sab automatically samajh aa jaayega.

Logarithm — definition as inverse of exponential

WHAT is a logarithm?

Deriving it from scratch (no memorising)

Worked examples

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Exponentials & Logarithms

Connections