log₆2 = a
2 = 6ᵃ
2 = (2·3)ᵃ
2 = 2ᵃ·3ᵃ
2/2ᵃ = 3ᵃ
2¹⁻ᵃ = 3ᵃ ①

①

log₅3 = b
3 = 5ᵇ
3ᵃ = (5ᵇ)ᵃ = 5ᵃᵇ ②

Sub ② into ①,

2¹⁻ᵃ = 5ᵃᵇ

2 = 5ᵃᵇ/⁽¹⁻ᵃ⁾

log₅2 = ab / (1 - a)

Alternatively,

log₆2 = a
log₂6 = 1/a
log₂(2·3) = 1/a
log₂2 + log₂3 = 1/a
1 + log₂3 = 1/a
log₂3 = 1/a - 1 = (1 - a)/a

1/log₂3 = a/(1 - a)


log₅3 = b

Change the base to 2,

log₂3 / log₂5 = b

log₂3 × log₅2 = b

log₅2 = b × 1/log₂3

log₅2 = b × a/(1 - a)

= ab/(1 - a)

Third way :

log₆2 = a

Change base to 5,

log₅2 / log₅6 = a
log₅2 = a log₅6 ①

Next,

log₅3 = b
log₅(6/2) = b
log₅6 - log₅2 = b
log₅2 + b = log₅6 ②


Sub ② into ① ,

log₅2 = a (log₅2 + b)
log₅2 = a log₅2 + ab
log₅2 - a log₅2 = ab
(1 - a) log₅2 = ab

log₅2 = ab/(1 - a)