Logarifmik ifodalarni shakl almashtirish: matematika darsi

$\log_6 4$ va $\log_6 9$ butun son emas, lekin ularning yig'indisi aniq $2$ ga teng, chunki $\log_6 4+\log_6 9=\log_6 36$ . Imtihon misollarida logarifmlar shunday xossalar bilan yo'qoladi. Har bir xossaning sharti bor va u ifodaning aniqlanish sohasini o'zgartirishi mumkin, shuning uchun bu darsda har formula o'z sharti bilan beriladi.

Asosiy logarifmik ayniyat

a^{\log_a b}=b

a>0,\ a\ne 1,\ b>0

\log_a b

asosni

b

ga aylantiradigan daraja, biz asosni aynan shu darajaga oshiramiz

\log_a a^c=c

a>0,\ a\ne 1,\ c\ \text{— istalgan son}

Teskari tomon

Birinchi tenglikning chap tomoni faqat $b>0$ da ma'noga ega: $2^{\log_2 x}=x$ tenglik $x>0$ da o'rinli, $x=-3$ da chap tomon mavjud emas. Ikkinchi tenglikda esa $c$ manfiy ham bo'lishi mumkin: $\log_2 2^{-3}=-3$ .

Ifoda	Hisob	Natija
Ifoda $5^{\log_5 7}$	Hisobayniyat	Natija $7$
Ifoda $3^{2\log_3 5}$	Hisob $3^{\log_3 25}$	Natija $25$
Ifoda $4^{\log_2 7}$	Hisob $2^{2\log_2 7}=2^{\log_2 49}$	Natija $49$
Ifoda $27^{\log_3 2}$	Hisob $3^{3\log_3 2}=3^{\log_3 8}$	Natija $8$
Ifoda $7^{1+\log_7 3}$	Hisob $7\cdot 7^{\log_7 3}=7\cdot 3$	Natija $21$
Ifoda $2^{3-\log_2 5}$	Hisob $\dfrac{2^3}{2^{\log_2 5}}=\dfrac{8}{5}$	Natija $\dfrac{8}{5}$
Ifoda $\left(\dfrac{1}{2}\right)^{\log_2 3}$	Hisob $2^{-\log_2 3}=\dfrac{1}{2^{\log_2 3}}$	Natija $\dfrac{1}{3}$
Ifoda $(\sqrt{3})^{2\log_3 5}$	Hisob $3^{\log_3 5}$	Natija $5$
Ifoda $5^{2\log_{25}6}$	Hisob $2\log_{25}6=\log_5 6$ , $5^{\log_5 6}$	Natija $6$
Ifoda $\left(1+4^{\log_2 3}\right)^{\lg 7}$	Hisob $(1+9)^{\lg 7}=10^{\lg 7}$	Natija $7$

Daraja asosi va logarifm asosi har xil bo'lsa, avval daraja asosini logarifm asosining darajasi ko'rinishida yozing.

Ko'paytmaning logarifmi

\log_a(bc)=\log_a b+\log_a c

a>0,\ a\ne 1,\ b>0,\ c>0

b=a^m

c=a^n

bo'lsa,

bc=a^{m+n}

: ko'paytirishda ko'rsatkichlar qo'shiladi

Formula istalgan sondagi musbat ko'paytuvchilar uchun o'rinli va ikki tomonga ishlaydi.

\log_6 4+\log_6 9=\log_6 36=2

Yig'indi bitta logarifmga keladi

\log_2 40=\log_2 8+\log_2 5

=3+\log_2 5

Logarifm yig'indiga yoyiladi. Shu kabi

\lg 20=1+\lg 2

Bo'linmaning logarifmi

\log_a\frac{b}{c}=\log_a b-\log_a c

a>0,\ a\ne 1,\ b>0,\ c>0

Bo'lishda ko'rsatkichlar ayiriladi. Xususan,

\log_a\dfrac{1}{c}=-\log_a c

$\log_2 40-\log_2 5=\log_2 8=3$ . $\lg 50=\lg\dfrac{100}{2}=2-\lg 2$ . $\log_3\dfrac{1}{81}=-\log_3 81=-4$ .

Darajaning logarifmi

\log_a b^k=k\log_a b

a>0,\ a\ne 1,\ b>0

k

istalgan son.

b=a^m

bo'lsa,

b^k=a^{mk}

\log_a\sqrt[n]{b}=\frac{1}{n}\log_a b

b>0,\ n\ \text{— natural},\ n\ge 2

Ildiz bu kasr ko'rsatkichli daraja

O'qilishiga e'tibor bering: $\log_a^2 b=(\log_a b)^2$ , $\log_a b^2=\log_a(b^2)=2\log_a b$ . $\log_2^2 8=3^2=9$ , $\log_2 8^2=\log_2 64=6$ .

Formula sohani o'zgartiradi

Yuqoridagi formulalar sonlar uchun xavfsiz. Ichida $x$ bo'lsa, chap va o'ng tomonning sohalari har xil bo'lishi mumkin. Chap tomonda ko'paytma musbat bo'lsa yetarli, o'ng tomonda esa har bir ko'paytuvchi musbat bo'lishi shart.

Ifoda	Sohasi	Ifoda	Sohasi
Ifoda $\log_a(xy)$	Sohasi $xy>0$	Ifoda $\log_a x+\log_a y$	Sohasi $x>0,\ y>0$
Ifoda $\log_a\dfrac{x}{y}$	Sohasi $\dfrac{x}{y}>0$	Ifoda $\log_a x-\log_a y$	Sohasi $x>0,\ y>0$
Ifoda $\log_a x^2$	Sohasi $x\ne 0$	Ifoda $2\log_a x$	Sohasi $x>0$
Ifoda $\log_a x^3$	Sohasi $x>0$	Ifoda $3\log_a x$	Sohasi $x>0$

Barcha holatlarda to'g'ri formulalar modul bilan yoziladi.

\log_a(xy)=\log_a|x|+\log_a|y|

xy>0

\log_a\frac{x}{y}=\log_a|x|-\log_a|y|

\dfrac{x}{y}>0

\log_a x^{2n}=2n\log_a|x|

x\ne 0

da,

n

natural

Son bilan	Hisob
$\log_2\big((-2)(-8)\big)$	$\log_2 16=4$ ; $\log_2(-2)+\log_2(-8)$ esa mavjud emas; to'g'ri yoyilmasi $\log_2 2+\log_2 8=1+3$
$\log_2(-4)^2$	$\log_2 16=4$ ; $2\log_2(-4)$ mavjud emas; to'g'risi $2\log_2\|-4\|=4$

Tenglamada bu shunday namoyon bo'ladi. Chapdan o'ngga o'tish, masalan $\lg x^2$ ni $2\lg x$ deb yozish, sohani toraytiradi va ildiz yo'qotadi: $\lg x^2=2$ ning ildizlari $x=\pm 10$ , $2\lg x=2$ ning ildizi faqat $x=10$ .

O'ngdan chapga o'tish, masalan $\log_3(x-1)+\log_3(x+1)$ ni $\log_3(x^2-1)$ deb yozish, sohani kengaytiradi va chet ildiz beradi:

\log_3(x-1)+\log_3(x+1)=1

x^2-1=3

x=\pm 2

, lekin

x=-2

chet ildiz, chunki

\log_3(-3)

mavjud emas. Javob:

x=2

Yig'indining logarifmi uchun formula yo'q

Noto'g'ri yozuv	Chap tomon	O'ng tomon
Noto'g'ri yozuv $\log_a(b+c)=\log_a b+\log_a c$	Chap tomon $\log_2(4+4)=3$	O'ng tomon $\log_2 4+\log_2 4=4$
Noto'g'ri yozuv $\log_a(b-c)=\log_a b-\log_a c$	Chap tomon $\log_2(16-8)=3$	O'ng tomon $\log_2 16-\log_2 8=1$
Noto'g'ri yozuv $\log_a b\cdot\log_a c=\log_a(bc)$	Chap tomon $\log_2 4\cdot\log_2 8=6$	O'ng tomon $\log_2 32=5$
Noto'g'ri yozuv $\dfrac{\log_a b}{\log_a c}=\log_a\dfrac{b}{c}$	Chap tomon $\dfrac{\log_2 64}{\log_2 4}=3$	O'ng tomon $\log_2\dfrac{64}{4}=4$

Ayrim sonlarda bunday tengliklar tasodifan bajarilishi mumkin, masalan $\log_2(2+2)=\log_2 2+\log_2 2$ , lekin bu umumiy qoida emas.

Yangi asosga o'tish

\log_a b=\frac{\log_c b}{\log_c a}

a>0,\ a\ne 1,\ b>0

c>0,\ c\ne 1

a^{\log_a b}=b

tenglikning ikkala tomonidan

c

asosli logarifm olamiz

$c$ son bo'lsa, soha o'zgarmaydi. Berilgan logarifmlar qaysi asosda bo'lsa, shu asosga o'tiladi. Masalan, $\log_8 32=\dfrac{\log_2 32}{\log_2 8}=\dfrac{5}{3}$ .

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