6.2-bo'lim : Logarifm funktsiyalari
18. \(\displaystyle \frac{1}{3}\log a - 6\log b + 2\) ni bitta koeffitsientli bitta logarifmaga birlashtiring.
Yechim
Buni bitta logarifmaga aylantirish uchun biz logarifmlarni teskari yo'nalishda ajratish uchun ishlatgan xususiyatlardan foydalanamiz. Ushbu jarayondagi birinchi qadam mulkdan foydalanish,
\[{\log _b}\left( {{x^r}} \right) = r{\log _b}x\]barcha logarifmlar bitta koeffitsientga ega ekanligiga ishonch hosil qilish. Buni birinchi navbatda qilish kerak, chunki logarifmlarning yig'indisini/farqlarini birlashtirishga imkon beradigan barcha xususiyatlar alohida logarifmlarda bitta koeffitsientni talab qiladi. Shunday qilib, ushbu xususiyatdan foydalanish,
\[\log \left( {{a^{\frac{1}{3}}}} \right) - \log \left( {{b^6}} \right) + 2\]2-qadam
Endi, 2-ga e'tibor bering, buni logarifm shaklida quyidagicha yozishimiz mumkin:
\[2 = \log {10^2} = \log 100\]E'tibor bering, bu haqiqatan ham mulkdan foydalanmoqda,
\[{\log _b}{b^x} = x\]Demak, endi bizda,
\[\log \left( {{a^{\frac{1}{3}}}} \right) - \log \left( {{b^6}} \right) + \log 100\]3-qadam
Endi bu nuqtadan chiqishning bir necha yo'li mavjud. Biz ikkita xususiyatdan birini ishlatishimiz mumkin.
\[{\log _b}\left( {xy} \right) = {\log _b}x + {\log _b}y\hspace{0.25in}\hspace{0.25in}{\log _b}\left( {\frac{x}{y}} \right) = {\log _b}x - {\log _b}y\]va aslida biz oxirida ikkalasini ham ishlatishimiz kerak bo'ladi.
Birinchi ikkita logarifm bir-biridan farq qiladi, shuning uchun keling, birinchi bo'lib olish uchun ularni birlashtirish uchun quotient xususiyatidan foydalanamiz,
\[\log \left( {{a^{\frac{1}{3}}}} \right) - \log \left( {{b^6}} \right) + \log {10^2} = \log \left( {\frac{{\sqrt[3]{a}}}{{{b^6}}}} \right) + \log 100\]Javobni biroz chiroyli qilish uchun biz birinchi atamadagi kasr ko'rsatkichini ildizga aylantirdik, lekin umuman buni qilish shart emas.
4-qadam
Va nihoyat, shuni yodda tutingki, bizda ikkita logarifm yig'indisi bor va biz ularni birlashtirish uchun mahsulot xususiyatidan foydalanishimiz mumkin,
\[\log \left( {{a^{\frac{1}{3}}}} \right) - \log \left( {{b^6}} \right) + \log 100 = \require{bbox} \bbox[2pt,border:1px solid black]{{\log \left( {\frac{{100\,\,\sqrt[3]{a}}}{{{b^6}}}} \right)}}\]