Logarithm — definition as inverse of exponential

3.2.6 · Maths › Exponentials & Logarithms

Logarithm KYA hota hai?

Restrictions KYUN hain?

$b > 0$ : negative bases jaise $(- 2)^{x}$ signs ke beech jump karte hain / complex ho jaate hain — yeh ek smooth function nahi hai.
$b \neq = 1$ : kyunki $1^{x} = 1$ hamesha hota hai, toh " $1$ ko kis power par raise karne se $5$ milega?" ka koi answer nahi — invertible nahi hai.
$y > 0$ : kyunki har real $x$ ke liye $b^{x} > 0$ hota hai, toh iska inverse sirf positive inputs hi accept kar sakta hai.

Isse scratch se derive karna (kuch bhi memorise mat karo)

Step 1. Exponential $y = b^{x}$ se shuru karo. Yeh step KYUN? Hume inverse chahiye, toh hum poochte hain ki kaunsa $x$ ek diye gaye $y$ se aaya.

Step 2. Roles swap karo: $y$ ke terms mein $x$ solve karo. $+, -, \times, \div$ use karke exponent se $x$ isolate karne ki koi algebra trick nahi hai — toh hum answer ke liye ek symbol invent karte hain aur isse $lo g_{b} y$ kehte hain. Yeh step KYUN? Exponent "upar phansa hua" hai; log woh tool hai jo specifically isse neeche laane ke liye banaya gaya hai.

Step 3. Definition ke anusar, $x = lo g_{b} y$ . $y = b^{x}$ mein wapas substitute karne par $y = b^{l o g_{b} y}$ milta hai — yeh hamari pehli golden identity hai. Yeh step KYUN? Substitution prove karta hai ki definition self-consistent hai.

Step 4. Iske bajaay $x$ se shuru karo: pehle $b^{(\cdot)}$ apply karo phir $lo g_{b}$ : $lo g_{b} (b^{x}) = x$ kyunki jo exponent $b^{x}$ produce karta hai woh literally $x$ hai. Yeh step KYUN? Confirm karta hai ki cancellation dono directions mein kaam karta hai ⇒ genuine inverse.

Figure — Logarithm — definition as inverse of exponential

Worked examples

Common mistake Common errors ka steel-manned analysis

Mistake A: " $lo g$ of a negative number theek hai, jaise $lo g_{2} (- 8) = - 3$ ." Kyun sahi lagta hai: students $- 3$ mein minus dekhte hain aur sochte hain ki yeh negative "banata" hai. Fix: $2^{- 3} = \frac{1}{8} > 0$ . Positive base ki koi bhi real power kabhi negative nahi hoti, toh $lo g_{2} (- 8)$ undefined hai. Exponent ka sign size control karta hai, output ka sign kabhi nahi.

Mistake B: " $lo g_{b} (x + y) = lo g_{b} x + lo g_{b} y$ ." Kyun sahi lagta hai: yeh un distributive/linear rules ki nakal karta hai jo hume pasand hain. Fix: logs multiplication ko addition mein convert karte hain, addition ko addition mein nahi. $lo g_{b} (x y) = lo g_{b} x + lo g_{b} y$ — lekin $lo g_{b} (x + y)$ ki koi simplification nahi hai. Test karo: $lo g_{10} (1 + 1) = lo g_{10} 2 \approx 0.30$ , jabki $lo g_{10} 1 + lo g_{10} 1 = 0$ . Equal nahi hain.

Mistake C: $lo g_{b} (b^{x}) = x$ ko $(lo g_{b} b)^{x}$ se confuse karna. Fix: $lo g_{b} (b^{x})$ matlab hai "woh exponent jo $b^{x}$ deta hai" $= x$ . $lo g_{b} b = 1$ , toh $(lo g_{b} b)^{x} = 1^{x} = 1$ . Bilkul alag hai.

Recall Feynman: ek 12-saal ke bachche ko explain karo

Ek magic doubling machine imagine karo. $1$ daalo, button $3$ baar dabao, aur yeh $1 \to 2 \to 4 \to 8$ tak badhta hai. Exponent yeh hai ki tune button kitni baar dabaya. Ab maano tumhare dost tumhe $8$ dete hain aur poochhhte hain, "Tune kitni baar dabaya?" Us sawaal ka jawab dena exactly wahi hai jo ek logarithm karta hai: yeh button-presses count karta hai. "Log base 2 of 8 = 3" ka matlab sirf yeh hai ki "tune doubling button 3 baar dabaya." Kyunki $1$ se doubling karke tum kabhi zero ya usse neeche nahi aa sakte, tum yeh sawaal sirf positive numbers ke baare mein pooch sakte ho.

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 $y=x$

lo g_{b} y

literally kya represent karta hai?

Woh exponent jis par

b

ko raise karne se

y

milta hai (yaani

b^{x} = y

ka solution

x

Log aur exponential form ko connect karne wali definition batao.

lo g_{b} y = x ⟺ b^{x} = y

, jahan

b > 0, b \neq = 1, y > 0

Base

b \neq = 1

KYUN hona chahiye?

Kyunki

1^{x} = 1

har

x

ke liye hota hai, toh

b^{x}

one-to-one nahi hai aur uska koi inverse nahi hai.

lo g_{b}

ka domain sirf

y > 0

KYUN hai?

Kyunki har real

x

ke liye

b^{x} > 0

hota hai, toh inverse sirf positive inputs hi accept kar sakta hai.

b^{l o g_{b} y}

simplify karo.

y

(dono operations cancel ho jaate hain; yeh

y > 0

ke liye valid hai).

lo g_{b} (b^{x})

simplify karo.

x

(saare real

x

ke liye).

lo g_{2} 8

evaluate karo.

3

, kyunki

2^{3} = 8

lo g_{3} \frac{1}{9}

evaluate karo.

- 2

, kyunki

3^{- 2} = \frac{1}{9}

y = b^{x}

ko kaun si line mein reflect karne se

y = lo g_{b} x

milta hai?

Line

y = x

Kya

lo g_{b} (x + y) = lo g_{b} x + lo g_{b} y

sach hai?

Nahi. Logs products ko sums mein convert karte hain:

lo g_{b} (x y) = lo g_{b} x + lo g_{b} y

; sums ke liye aisa koi rule nahi hai.

Har

lo g_{b} x

graph kaun se point se guzarta hai, aur kyun?

(1, 0)

se, kyunki

b^{0} = 1

hai toh

lo g_{b} 1 = 0

y = lo g_{b} x

ka vertical asymptote kya hai?

x = 0

(

b^{x}

ke horizontal asymptote

y = 0

ka image).