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answers:
la console say: Log[x](40) = p → you know that: Log[a](x) = Ln(x)/Ln(a) where a is the base
Ln(40) / Ln(x) = p
Ln(x) = (1/p).Ln(40) ← memorize this result as (1)
Log[x](50) = q → you know that: Log[a](x) = Ln(x)/Ln(a) where a is the base
Ln(50) / Ln(x) = q
Ln(x) = (1/q).Ln(50) ← memorize this result as (2)
Ln(x) = (1/q).Ln(50) → recall (1): Ln(x) = (1/p).Ln(40)
(1/p).Ln(40) = (1/q).Ln(50)
q/p = Ln(50) / Ln(40)
q/p = Ln(25 * 2) / Ln(8 * 5) → you know that: Ln(ab) = Ln(a) + Ln(b)
q/p = [Ln(25) + Ln(2)] / [Ln(8) + Ln(5)] → you know that: Ln(25) = Ln(5^2) = 2.Ln(5)
q/p = [2.Ln(5) + Ln(2)] / [Ln(8) + Ln(5)] → you know that: Ln(8) = Ln(2^3) = 3.Ln(2)
q/p = [2.Ln(5) + Ln(2)] / [3.Ln(2) + Ln(5)]
q.[3.Ln(2) + Ln(5)] = p.[2.Ln(5) + Ln(2)]
3q.Ln(2) + q.Ln(5) = 2p.Ln(5) + p.Ln(2)
3q.Ln(2) - p.Ln(2) = 2p.Ln(5) - q.Ln(5)
(3q - p).Ln(2) = (2p - q).Ln(5)
Ln(2) / Ln(5) = (2p - q)/(3q - p) ← memorize this result as (3)
Ln(5) / Ln(2) = (3q - p)/(2p - q) ← memorize this result as (4)
You restart from (1)
Ln(x) = (1/p).Ln(40) → you divide by Ln(2) both sides
Ln(x) / Ln(2) = (1/p).Ln(40) / Ln(2)
Log[2](x) = (1/p).Ln(40) / Ln(2) → you know that: Ln(40) = Ln(8) + Ln(5) = 3.Ln(2) + Ln(5)
Log[2](x) = (1/p).[3.Ln(2) + Ln(5)] / Ln(2)
Log[2](x) = (1/p).[3 + {Ln(5) / Ln(2)}] → recall (4): Ln(5) / Ln(2) = (3q - p)/(2p - q)
Log[2](x) = (1/p).[3 + {(3q - p)/(2p - q)}]
Log[2](x) = (1/p).[{3.(2p - q) + (3q - p)}/(2p - q)]
Log[2](x) = (1/p).[{6p - 3q + 3q - p}/(2p - q)]
Log[2](x) = (1/p).[5p/(2p - q)]
Log[2](x) = 5/(2p - q)
You restart from (2)