Question

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Answer 1

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.)

_________________________

Answer 2

$\rule{200}2$

$\huge\tt{GIVEN:}$

• If log_12 27 = a

$\rule{200}2$

$\huge\tt{TO~FIND:}$

• log_6 16

$\rule{200}2$

$\huge\tt{SOLUTION:}$

↪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)

$\rule{200}1$

↪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)

↪y = (12- 4a) / (3 + a)

$\rule{200}2$