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Answers
Qᴜᴇsᴛɪᴏɴ :-
if log₁₂27 = a, then find log₆16 = y ?
Fᴏʀᴍᴜʟᴀ ᴜsᴇᴅ :-
- log(a)^b = b*log(a)
- log(a * b) = log(a) + log(b)
- Log(b)x = Log(a)x/Log(a)b
- log(a)a = 1
- logb(c) = 1 / logc(b)
Sᴏʟᴜᴛɪᴏɴ :-
→ log₁₂27 = a
→ log₁₂(3)³ = a
using log(a)^b = b*log(a) in LHS now,
→ 3*log₁₂(3) = a
Now, using log changing base formula , Log(b)x = Log(a)x/Log(a)b in LHS, we get,
→ [ (3*log₃3) / (log₃12) ] = a
Now in Numerator , using same base formula , ( = 1) ,
→ ( 3 * 1) / (log₃12) = a
→ a = 3/(log₃12)
→ a = 3/log₃(3*4)
→ a = 3/log₃(3*2²)
Again using log(a * b) = log(a) + log(b) in Denominator , we get,
→ a = 3/[ log₃(3) + log₃(2)² ]
using log(a)^b = b*log(a)
→ a = 3/[ log₃(3) + 2*log₃(2) ]
using log(a)a = 1 Now,
→ a = 3/[ 1 + 2*log₃(2) ]
→ [ 1 + 2*log₃(2) ] = (3/a)
→ 2*log₃(2) = (3/a) - 1
→ log₃(2) = (1/2) [ (3/a) - 1 ]
→ log₃(2) = (1/2) [ (3 - a)/a ]
→ log₃(2) = (3 - a)/2a
Now, using Logarithm Switch Rule logb(c) = 1 / logc(b) we get,
→ log₂(3) = 1/[(3 - a)/2a]
→ log₂(3) = 2a / (3 - a) ---------- Equation ❶
_________________
Now, Solving Finding Part :-
→ y = log₆16
→ y = log₆(2)⁴
using log(a)^b = b*log(a)
→ y = 4 * log₆(2)
Now, using logb(c) = 1 / logc(b)
→ y = 4/log₂(6)
→ y = 4 / log₂(2*3)
using log(a * b) = log(a) + log(b) in Denominator,
→ y = 4 / [ log₂(2) + log₂(3) ]
using log(a)a = 1 Now,
→ y = 4 / [ 1 + log₂(3) ]
Now, Putting value of Equation ❶ Here, we get,
→ y = 4 / [ 1 + {2a/(3 - a)} ]
→ y = 4 / [ {3 - a + 2a} / (3 - a) ]
→ y = 4 / [ (3 + a) / (3 - a) ]
→ y = 4(3 - a) / (3 + a)
→ y = (12 - 4a) / (3 + a) (Ans.)
_________________________
- If log_12 27 = a
- log_6 16
↪log_12 27 = a
↪log_12 (3)³ = a
↪3 × log_12 (3) = a
↪[(3×log_3 3) / (log_3 12) ] = a
↪(3×1)/(log_3 12) = a
↪a = 3/log_3(3×2²)
↪a = 3/[log_3 (3) + log_3 (2)²]
↪a = 3/[1+2×log_3(2)]
↪[1+2 × log_3 (2) ] = (3/a)
↪2 × log_3 (2) = (3/a)
↪log_3 (2) = 1/2 [(3-a)/2a
↪log_2 (3) = 1/[(3-a)/2a]
↪log_2 (3) = 2a /(3-a) __(EQ.1)
↪y = log_6 (16)
↪y = 4 × log_6 (2)
↪y = 4/log_2 (2×3)
↪y = 4/ [log_2(2) + log_2(3) ]
↪y = 4/[1+log_2(3)]
↪y = 4/[1+{2a/(3-a)}]
↪y = 4/[{3-a+2a}/(3-a)]
↪y = (3-a)/(3+a)