1.9-bo'lim : Eksponensial va logarifm tenglamalari
5. \(2{{\bf{e}}^{3\,y + 8}} - 11{{\bf{e}}^{5 - 10\,y}} = 0\) ning barcha yechimlarini toping. Agar yechimlar bo'lmasa, sababini aniq tushuntiring.
Maslahat: Bu erda davom etishning eng yaxshi usuli - tenglamani bitta eksponensialga kamaytirishdir.
Yechim
Tenglamadagi ikkala eksponensial bilan buni echish biroz qiyin bo'lishi mumkin, shuning uchun keling, buni bitta eksponensial tenglamaga tushirish uchun biroz ish qilaylik.
\[\begin{align*}2{{\bf{e}}^{3\,y + 8}} & = 11{{\bf{e}}^{5 - 10\,y}}\\ \frac{{{{\bf{e}}^{3\,y + 8}}}}{{{{\bf{e}}^{5 - 10\,y}}}} & = \frac{{11}}{2}\\ {{\bf{e}}^{13\,y + 3}} & = \frac{{11}}{2}\end{align*}\]E'tibor bering, biz eksponensialga bo'linishimiz mumkin edi, lekin biz qilgan ko'rsatkichga bo'lish orqali biz \(y\) da salbiy ko'rsatkichdan qochamiz, ba'zida uni yo'qotish oson.
2-qadam
Endi biz qilishimiz kerak bo'lgan narsa ikkala tomonning logarifmini olish va \(y\) ni hal qilishdir.
\[\begin{align*}\ln \left( {{{\bf{e}}^{13\,y + 3}}} \right) & = \ln \left( {\frac{{11}}{2}} \right)\\ 13y + 3 & = \ln \left( {\frac{{11}}{2}} \right)\\ y & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{13}}\left( {\ln \left( {\frac{{11}}{2}} \right) - 3} \right) = - 0.09963}}\end{align*}\]Sizning afzalliklaringizga qarab, aniq yoki o'nlik echimdan foydalanish mumkin.