Exercise 1.4 — Logarithmic aur Exponential Equations/Inequalities Solve Karna
Bunyadi Concept
Is exercise mein do cheezein seekhni hain: equations solve karna aur inequalities solve karna, jab log ya exponential functions involved hon.
Sabse Zaroori Rule (Equations ke liye): Agar logₐ(A) = logₐ(B) (same base ho), to seedha A = B likh sakte hain (log hata sakte hain). Lekin hamesha domain check karna zaroori hai — kyunke log sirf positive numbers ka hota hai, isliye jo answer aaye usay original equation ke domain mein check karna parta hai (warna galat root include ho jayega).
Part 1 — Equations (Question 1)
(i) log(3x−2) = log(x²−5x+6)
- Same base (log₁₀ dono taraf) hai, isliye arguments barabar kar diye: 3x−2 = x²−5x+6
- Quadratic ban gayi: x²−8x+8=0, quadratic formula se do roots mile: x = 4±2√2
- Domain check kiya (dono arguments positive hone chahiye) — dono roots satisfy karte hain
- Dono answers valid hain
(ii) ln(x²−1) = ln(3x−2)
- Arguments barabar: x²−1 = 3x−2 → x²−3x+1=0
- Do roots mile: x = (3±√5)/2
- Yahan important step hai: domain check karne par pata chala ke ek root (≈0.38) domain fail karta hai (kyunke x²−1 negative ho jata hai), isliye wo reject ho gaya
- Sirf ek answer valid hai: x=(3+√5)/2
Yahan se sabak: Har log equation solve karne ke baad domain zaroor check karo — kabhi ek root reject ho jata hai.
(iii) log(x) + log(x²−5x+7) = log(3)
- Rule use kiya:
log A + log B = log(AB), isliye left side combine ho gaya - Cubic equation bani: x³−5x²+7x−3=0
- Trial method se x=1 check kiya, satisfy hua, isliye (x−1) factor nikla
- Baaki quadratic ko factorize kiya: (x−1)(x−1)(x−3)=0, yani x=1 (do dafa) aur x=3
- Domain check kiya, dono valid nikle
(iv) 3^(x+1) − 2·9ˣ + 9 = 0
- Trick: 3^(x+1) ko 3·3ˣ likha, aur 9ˣ ko (3ˣ)² likha
- Phir substitution kiya: y = 3ˣ, equation ban gayi quadratic mein: 2y²−3y−9=0
- Quadratic formula se do y values mile: y=3 ya y=−1.5
- Chunke 3ˣ hamesha positive hota hai, isliye y=−1.5 reject
- 3ˣ=3 se x=1 mila
(v) x·log_(1/e)4 = log_(1/e)(16³−3)
- Rule use kiya:
n·log A = log(Aⁿ), isliye left side ko log_(1/e)(4ˣ) likha - Same base dono taraf, isliye arguments barabar: 4ˣ = 16³−3 = 4093
- Dono taraf log liya (base 4 mein): x = log₄(4093) ≈ 5.9998
(vi) 2ˣ + 2⁻ˣ = 5
- Substitution: y=2ˣ, note kiya ke 2⁻ˣ = 1/y
- Equation bani: y + 1/y = 5 → y²−5y+1=0
- Quadratic formula se: y = (5±√21)/2
- Phir dono taraf log₂ liya kyunke y=2ˣ tha
- Do answers mile jo ek dusre ke negative hain
Part 2 — Inequalities (Question 2)
Important Rule Yaad Rakho: Jab log ya exponential inequality solve karte hain:
- Agar base > 1 ho, to inequality ki direction same rehti hai jab log hataate hain
- Agar base < 1 ho (jaise 1/3), to inequality ki direction ulti (reverse) ho jaati hai
- Hamesha domain pehle nikalo, phir answer ko domain ke saath intersect karo
(i) 2ˣ + 2¹⁻ˣ ≤ 3
- 2¹⁻ˣ ko 2/2ˣ likha, phir y=2ˣ substitute kiya
- Inequality bani: y + 2/y ≤ 3 → (y−1)(y−2) ≤ 0
- Ye solve hua: 1≤y≤2, matlab 1≤2ˣ≤2
- Answer: 0 ≤ x ≤ 1
(ii) 2ˣ ≥ log₂(16)
- Pehle log₂(16)=4 nikala (kyunke 2⁴=16)
- 2ˣ≥4=2² bana, base 2>1 hai isliye direction same rahi
- Answer: x≥2
(iii) log₂(x−1) ≥ log₂(5−x)
- Pehle domain nikala: x>1 aur x<5, yani (1,5)
- Base 2>1 hai, isliye direction preserve hui: x−1≥5−x → x≥3
- Domain ke saath intersect kiya
- Answer: 3≤x<5
(iv) log_(1/3)(x²−4x+3) > log_(1/3)(2x−1)
- Domain nikala (dono arguments positive hone chahiye): (1/2,1)∪(3,∞)
- Base 1/3 < 1 hai, isliye direction REVERSE ho gayi jab log hataya
- x²−4x+3 < 2x−1 → quadratic solve karke: (3−√5, 3+√5)
- Domain ke saath intersect kiya
- Answer: x∈(3−√5,1)∪(3,3+√5)
(v) log₂(x−2)+log₂(x−3) ≥ 3
- Domain: x>3
- Logs combine kiye: log₂[(x−2)(x−3)]≥3, phir exponential form mein liya: (x−2)(x−3)≥2³=8
- Quadratic solve kiya, roots mile (5±√33)/2
- Domain ke saath intersect kiya
- Answer: x ≥ (5+√33)/2
(vi) log(x)+log(x−3) ≥ log(2x)+log(x−1)
- Domain: x>3 (sabse strict condition)
- Dono taraf logs combine kiye, base 10>1 isliye direction same
- Simplify karne par: x(x+1)≤0 → −1≤x≤0
- Ye range domain (x>3) se bilkul match nahi karti
- Answer: No solution exists
Sabak: Kabhi kabhi jo answer equation solve karne se milta hai, wo diye gaye domain ke saath overlap hi nahi karta — is soorat mein answer “no solution” hota hai.
