10=ln((t^2 1/2)+e)

解

10 = ln ((t^{2} \frac{1}{2}) + e)

t = 2 e^{10} - 2 e, t = - 2 e^{10} - 2 e

十進法表記

t = 209.87495 \dots, t = - 209.87495 \dots

解答ステップ

10 = ln ((t^{2} \frac{1}{2}) + e)

対数の規則を適用する

e^{10} = t^{2} \frac{1}{2} + e

解く

e^{10} = t^{2} \frac{1}{2} + e : t = 2 e^{10} - 2 e, t = - 2 e^{10} - 2 e

t = 2 e^{10} - 2 e, t = - 2 e^{10} - 2 e