ln(t-m)=kt+c

Lösung

ln (t - m) = k t + c

t = - \frac{W ( - \frac{k}{e ^{- mk - c}} ) - mk}{k}

Schritte zur Lösung

ln (t - m) = k t + c

Wenn

f (x) = g (x),

dann

a^{f (x)} = a^{g (x)}

e^{l n (t - m)} = e^{k t + c}

Vereinfache

e^{l n (t - m)} : t - m

t - m = e^{k t + c}

Löse

t - m = e^{k t + c} : t = - \frac{W ( - \frac{k}{e ^{- mk - c}} ) - mk}{k}

t = - \frac{W ( - \frac{k}{e ^{- mk - c}} ) - mk}{k}

lo g_{10} (x^{2}) = - 10

so l v e f or y, ln (y - 1) - 2 = x + ln (x)

ln (x + 2) = ln (7 x)

ln (3 x + 6) = 5

ln (1 - x) + ln (3 - x) = ln (8)