6.4-bo'lim : Logarifm tenglamalarini yechish
5. Quyidagi tenglamani yeching.
\[{\log _2}\left( {x + 1} \right) - {\log _2}\left( {2 - x} \right) = 3\]Yechim
Avval shuni ta'kidlaymizki, biz chap tomondagi ikkita logarifmni birlashtira olamiz:
\[{\log _2}\left( {\frac{{x + 1}}{{2 - x}}} \right) = 3\]2-qadam
Endi biz buni osongina eksponensial shaklga o'tkazishimiz mumkin.
\[\frac{{x + 1}}{{2 - x}} = {2^3} = 8\]3-qadam
Endi biz qilishimiz kerak bo'lgan yagona narsa 2-bosqichdagi tenglamani yechishdir va bu biz qanday echishni biladigan tenglama. Mana yechim ishi.
\[\begin{align*}\frac{{x + 1}}{{2 - x}} & = 8\\ x + 1 & = 8\left( {2 - x} \right) = 16 - 8x\\ 9x & = 15\hspace{0.25in} \to \hspace{0.25in}x = \frac{{15}}{9} = \frac{5}{3}\end{align*}\]4-qadam
Yakuniy qadam sifatida biz yuqoridagi bosqichdagi raqamni olishimiz va nol yoki manfiy sonlarning logarifmini qabul qilmasligimiz uchun uni muammo bayonotidagi asl tenglamaga ulashimiz kerak!
Mana raqamni tekshirish ishi.
\[x = \frac{5}{3}:\] \[\begin{align*}{\log _2}\left( {\frac{5}{3} + 1} \right) - {\log _2}\left( {2 - \frac{5}{3}} \right) & = 3\\ {\log _2}\left( {\frac{8}{3}} \right) - {\log _2}\left( {\frac{1}{3}} \right) & = 3\hspace{0.25in}\,\,\,\,\,{\mbox{OKAY}}\end{align*}\]
Bunday holda, raqam logarifmlarda manfiy sonlarni keltirmadi, shuning uchun bu aslida yechim (har bir muammo bilan bo'lmaydi, shuning uchun har doim ham shunday bo'lishini kutmang!).
Demak, tenglamaning yechimi u holda: \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = \frac{5}{3}}}\).