Question

задача с логарифмами во вложении

Answer 1

log(a) b = 1/ log(b) a

log(a) b * log(b) a = 1

log (a^n) b^k = k/n * log(a) b

log(a) a = 1

log(2) 3 * log(3) 2 - log(4) ∛2 = 1 - log(2^2) 2 ^1/3 = 1 - 1/3 * 1/2 * log(2) 2 = 1 - 1/6 = 5/6

Answer 2

㏒₂3*㏒₃2-㏒₄∛2=㏒₂3*(㏒₂2/㏒₂3) - ㏒₂²2¹/³=㏒₂3*(1/㏒₂3) -(1/6)㏒₂2=

1-(1/6)*1=5/6

использовал правило перехода к новому основанию

㏒₂5= ㏒₅5/㏒₅2=1//㏒₅2

и ㏒ₐˣсᵇ=(b/x)*㏒ₐIсI