4.5: Mali ya Logarithmic

Malengo ya kujifunza

• Tumia utawala wa bidhaa kwa logarithms.
• Tumia utawala wa quotient kwa logarithms.
• Tumia utawala wa nguvu kwa logarithms.
• Panua maneno ya logarithmic.
• Condense maneno ya logarithmic.
• Tumia formula ya mabadiliko-ya msingi kwa logarithms.

Katika kemia, kiwango cha pH kinatumika kama kipimo cha asidi au alkalinity ya dutu. Vipengele$7$ vilivyo na pH chini ya kuchukuliwa kuwa tindikali, na vitu vyenye pH kubwa zaidi kuliko$7$ inasemekana kuwa alkali. Miili yetu, kwa mfano, inapaswa kudumisha pH karibu na$7.35$ ili enzymes kufanya kazi vizuri. Ili kujisikia kwa nini ni tindikali na ni nini alkali, fikiria viwango vya pH zifuatazo za vitu vingine vya kawaida:

• Asidi ya betri:$0.8$
• Asidi ya tumbo:$2.7$
• Juisi ya machungwa:$3.3$
• Maji safi:$7$ katika$25^\circ C$
• Damu ya binadamu:$7.35$
• Nazi safi:$7.8$
• Hidroksidi ya sodiamu (lye):$14$

Kuamua kama suluhisho ni tindikali au alkali, tunapata pH yake, ambayo ni kipimo cha idadi ya ions hidrojeni chanya katika suluhisho. PH inaelezwa na formula ifuatayo, wapi$[\ce{H^{+}}]$ mkusanyiko wa ions hidrojeni katika suluhisho

$$\begin{align} {pH}&=−{\log}([\ce{H^{+}}]) \label{eq1} \\[4pt] &={\log}\left(\dfrac{1}{[\ce{H^{+}}]}\right) \label{eq2} \end{align}$$

Ulinganisho wa Ulinganisho\ ref {eq1} na\ ref {eq2} ni mojawapo ya mali ya logarithm tutakayochunguza katika sehemu hii.

Kielelezo$\PageIndex{1}$: PH ya asidi hidrokloric inajaribiwa na karatasi ya litmus. (mikopo: David Berardan).

Kutumia Kanuni ya Bidhaa kwa Logarithms

Kumbuka kwamba kazi za logarithmic na za kielelezo “tengeneze” kila mmoja. Hii ina maana kwamba logarithms zina mali sawa na watazamaji. Baadhi ya mali muhimu ya logarithms hutolewa hapa. Kwanza, mali zifuatazo ni rahisi kuthibitisha.

$$\begin{align*} \log_b1 &=0 \\[4pt] \log_bb &=1 \end{align*}$$

Kwa mfano,${\log}_51=0$ tangu$5^0=1$. Na${\log}_55=1$ tangu$5^1=5$.

Kisha, tuna mali inverse.

$$\begin{align*} \log_b(b^x) &=x \\[4pt] b^{\log_b x} &=x, \,x>0 \end{align*}$$

Kwa mfano, kutathmini${\log(100)}$, tunaweza kuandika upya logarithm kama${\log}_{10}({10}^2)$, na kisha kutumia mali inverse${\log}_b(b^x)=x$ kupata${\log}_{10}({10}^2)=2$.

Kutathmini$e^{\ln(7)}$, tunaweza kuandika upya logarithm kama$e^{{\log}_e7}$, na kisha kutumia mali inverse$b^{{\log}_bx}=x$ kupata$e^{{\log}_e 7}=7$.

Hatimaye, tuna mali moja kwa moja.

$$\log_bM = \log_bN \text{ if and only if } M=N$$

Tunaweza kutumia mali moja kwa moja kutatua equation${\log}_3(3x)={\log}_3(2x+5)$ kwa$x$. Kwa kuwa besi ni sawa, tunaweza kutumia mali moja kwa moja kwa kuweka hoja sawa na kutatua kwa$x$:

$3x=2x+5$Weka hoja sawa.

$x=5$Ondoa$2x$.

Lakini vipi kuhusu equation${\log}_3(3x)+{\log}_3(2x+5)=2$? Mali moja kwa moja haitusaidia katika mfano huu. Kabla ya kutatua equation kama hii, tunahitaji njia ya kuchanganya maneno upande wa kushoto wa equation.

Kumbuka kwamba sisi kutumia utawala wa bidhaa ya exponents kuchanganya bidhaa ya exponents kwa kuongeza:$x^ax^b=x^{a+b}$. Tuna mali sawa kwa logarithms, inayoitwa utawala wa bidhaa kwa logarithms, ambayo inasema kwamba logarithm ya bidhaa ni sawa na jumla ya logarithms. Kwa sababu magogo ni exponents, na sisi kuzidisha kama besi, tunaweza kuongeza exponents. Tutatumia mali inverse kupata utawala wa bidhaa hapa chini.

Kutokana$x$ na idadi yoyote halisi na chanya idadi halisi$M$$N$,$b$, na$b≠1$, ambapo, sisi kuonyesha

${\log}_b(MN)={\log}_b(M)+{\log}_b(N)$.

Hebu$m={\log}_bM$ na$n={\log}_bN$. Katika fomu ya kielelezo, equations hizi ni$b^m=M$ na$b^n=N$. Inafuata kwamba

$$\begin{align*} {\log}_b(MN)&= {\log}_b(b^mb^n) \qquad \text{Substitute for M and N}\\[4pt] &= {\log}_b(b^{m+n}) \qquad \text{Apply the product rule for exponents}\\[4pt] &= m+n \qquad \text{Apply the inverse property of logs}\\[4pt] &= {\log}_b(M)+{\log}_b(N) \qquad \text{Substitute for m and n} \end{align*}$$

Kumbuka kwamba matumizi ya mara kwa mara ya utawala wa bidhaa kwa logarithms inatuwezesha kurahisisha logarithm ya bidhaa ya mambo yoyote. Kwa mfano, fikiria${\log}_b(wxyz)$. Kutumia utawala wa bidhaa kwa logarithms, tunaweza kuandika upya logarithm hii ya bidhaa kama jumla ya logarithms ya mambo yake:

${\log}_b(wxyz)={\log}_bw+{\log}_bx+{\log}_by+{\log}_bz$

Utawala wa Bidhaa kwa Logarithms

Utawala wa bidhaa kwa logarithms unaweza kutumika kurahisisha logarithm ya bidhaa kwa kuandika upya kama jumla ya logarithms ya mtu binafsi.

$$\begin{align} {\log}_b(MN)={\log}_b(M)+{\log}_b(N)\text{ for } b> 0 \end{align}$$

Jinsi ya: Kutokana na logarithm ya bidhaa, tumia utawala wa bidhaa wa logarithms kuandika jumla sawa ya logarithms

1. Fanya hoja kabisa, akielezea kila sababu ya namba nzima kama bidhaa ya primes.
2. Andika maneno sawa kwa kuhesabu logarithms ya kila sababu.

Mfano$\PageIndex{1}$: Using the Product Rule for Logarithms

Panua${\log}_3(30x(3x+4))$.

Suluhisho

Tunaanza kwa kuzingatia hoja kabisa, kuonyesha$30$ kama bidhaa ya primes.

${\log}_3(30x(3x+4))={\log}_3(2⋅3⋅5⋅x⋅(3x+4))$

Halafu tunaandika equation sawa kwa kuhesabu logarithms ya kila sababu.

${\log}_3(30x(3x+4))={\log}_3(2)+{\log}_3(3)+{\log}_3(5)+{\log}_3(x)+{\log}_3(3x+4)$

Zoezi$\PageIndex{1}$

Panua${\log}_b(8k)$.

Jibu

${\log}_b2+{\log}_b2+{\log}_b2+{\log}_bk=3{\log}_b2+{\log}_bk$

Kutumia Utawala wa Quotient kwa Logarithms

Kwa quotients, tuna kanuni sawa ya logarithms. Kumbuka kwamba sisi kutumia utawala quotient ya exponents kuchanganya quotient ya exponents kwa kutoa:$x^{\frac{a}{b}}=x^{a−b}$. Utawala wa quotient kwa logarithms unasema kwamba logarithm ya quotient ni sawa na tofauti ya logarithms.

Utawala wa Quotient kwa Logarithms

Utawala wa quotient kwa logarithms unaweza kutumika kurahisisha logarithm au quotient kwa kuandika upya kama tofauti ya logarithms ya mtu binafsi.

$${\log}_b\left(\dfrac{M}{N}\right)={\log}_bM−{\log}_bN$$

Kama vile utawala wa bidhaa, tunaweza kutumia mali inverse kupata utawala quotient.

Ushahidi

Kutokana$x$ na idadi yoyote halisi na chanya idadi halisi$M$$N$,, na b, b$b≠1$, ambapo, sisi kuonyesha

${\log}_b\left(\dfrac{M}{N}\right)={\log}_b(M)−{\log}_b(N)$.

Hebu$m={\log}_bM$ na$n={\log}_bN$. Katika fomu ya kielelezo, equations hizi ni$b^m=M$ na$b^n=N$. Inafuata kwamba

$$\begin{align*} {\log}_b\left (\dfrac{M}{N} \right )&= {\log}_b\left(\dfrac{b^m}{b^n}\right) \qquad \text{Substitute for M and N}\\[4pt] &= {\log}_b(b^{m-n}) \qquad \text{Apply the quotient rule for exponents}\\[4pt] &= m-n \qquad \text{Apply the inverse property of logs}\\[4pt] &= {\log}_b(M)-{\log}_b(N) \qquad \text{Substitute for m and n} \end{align*}$$

Kwa mfano, kupanua${\log}\left (\dfrac{2x^2+6x}{3x+9} \right )$, lazima kwanza tueleze quotient kwa maneno ya chini kabisa. Factoring na kufuta sisi kupata,

$$\begin{align*} {\log}\left (\dfrac{2x^2+6x}{3x+9} \right )&= {\log}\left (\dfrac{2x(x+3)}{3(x+3)} \right ) \qquad \text{Factor the numerator and denominator}\\[4pt] &= {\log}\left (\dfrac{2x}{3} \right ) \qquad \text{Cancel the common factors} \end{align*}$$

Halafu tunatumia utawala wa quotient kwa kuondoa logarithm ya denominator kutoka kwa logarithm ya nambari. Kisha tunatumia utawala wa bidhaa.

$$\begin{align*} {\log}\left(\dfrac{2x}{3}\right) &={\log}(2x)−{\log}(3) \\[4pt] &={\log}(2)+{\log}(x)−{\log}(3) \end{align*}$$

Jinsi ya: Kutokana na logarithm ya quotient, tumia utawala wa quotient wa logarithms kuandika tofauti sawa ya logarithms

1. Eleza hoja kwa maneno ya chini kwa kuzingatia namba na denominator na kufuta maneno ya kawaida.
2. Andika maneno sawa kwa kuondoa logarithm ya denominator kutoka kwa logarithm ya nambari.
3. Angalia ili kuona kwamba kila neno limeenea kikamilifu. Ikiwa sio, tumia utawala wa bidhaa kwa logarithms kupanua kabisa.

Mfano$\PageIndex{2}$: Using the Quotient Rule for Logarithms

Panua${log}_2\left(\dfrac{15x(x−1)}{(3x+4)(2−x)}\right)$.

Suluhisho

Kwanza tunaona kwamba quotient ni sababu na kwa maneno ya chini, kwa hiyo tunatumia utawala wa quotient.

${\log}_2(\dfrac{15x(x−1)}{(3x+4)(2−x)})={\log}_2(15x(x−1))−{\log}_2((3x+4)(2−x))$

Angalia kwamba maneno yanayotokana ni logarithms ya bidhaa. Ili kupanua kabisa, tunatumia utawala wa bidhaa, akibainisha kuwa sababu kuu za sababu 15 ni 3 na 5.

$$\begin{align*} {\log}_2(15x(x-1))-{\log}_2((3x+4)(2-x))&= [{\log}_2(3)+{\log}_2(5)+{\log}_2(x)+{\log}_2(x-1)]-[{\log}_2(3x+4)+{\log}_2(2-x)]\\[4pt] &= {\log}_2(3)+{\log}_2(5)+{\log}_2(x)+{\log}_2(x-1)-{\log}_2(3x+4)-{\log}_2(2-x) \end{align*}$$

Uchambuzi

Kuna tofauti za kuzingatia katika mifano hii na baadaye. Kwanza, kwa sababu denominators lazima kamwe kuwa sifuri, maneno haya hayafafanuliwa kwa$x=−\dfrac{4}{3}$ na$x=2$. Pia, tangu hoja ya logarithm lazima iwe chanya, tunaona tunapoona logarithm iliyopanuliwa$x>0$,$x>1$ hiyo,$x>−\dfrac{4}{3}$, na$x<2$. Kuchanganya hali hizi ni zaidi ya upeo wa sehemu hii, na hatuwezi kuzingatia hapa au katika mazoezi yafuatayo.

Zoezi$\PageIndex{2}$

Panua${\log}_3\left(\dfrac{7x^2+21x}{7x(x−1)(x−2)}\right)$.

Jibu

${\log}_3(x+3)−{\log}_3(x−1)−{\log}_3(x−2)$

Kutumia Utawala wa Nguvu kwa Logarithms

Tumechunguza utawala wa bidhaa na utawala wa quotient, lakini tunawezaje kuchukua logarithm ya nguvu, kama vile$x^2$? Njia moja ni kama ifuatavyo:

$$\begin{align*} {\log}_b(x^2)&= {\log}_b(x\cdot x)\\[4pt] &= {\log}_bx+{\log}_bx\\[4pt] &= 2{\log}_bx \end{align*}$$

Angalia kwamba tulitumia utawala wa bidhaa kwa logarithms ili kupata suluhisho kwa mfano hapo juu. Kwa kufanya hivyo, tumepata utawala wa nguvu kwa logarithms, ambayo inasema kwamba logi ya nguvu ni sawa na mara exponent logi ya msingi. Kumbuka kwamba, ingawa pembejeo ya logarithm haiwezi kuandikwa kama nguvu, tunaweza kuibadilisha kwa nguvu. Kwa mfano,

$100={10}^2$$\sqrt{3}=3^{\dfrac{1}{2}}$$\dfrac{1}{e}=e^{−1}$

Utawala wa Nguvu kwa Logarithms

Utawala wa nguvu kwa logarithms unaweza kutumika kurahisisha logarithm ya nguvu kwa kuandika tena kama bidhaa ya mara exponent logarithm ya msingi.

$${\log}_b(M^n)=n{\log}_bM$$

Jinsi ya: Kutokana na logarithm ya nguvu, tumia utawala wa nguvu wa logarithms kuandika bidhaa sawa ya sababu na logarithm

1. Eleza hoja kama nguvu, ikiwa inahitajika.
2. Andika kujieleza sawa kwa kuzidisha mara exponent logarithm ya msingi.

Mfano$\PageIndex{3}$: Expanding a Logarithm with Powers

Panua${\log}_2x^5$.

Suluhisho

Hoja tayari imeandikwa kama nguvu, hivyo sisi kutambua exponent, 5, na msingi,$x$, na kuandika upya kujieleza sawa na kuzidisha mara exponent logarithm ya msingi.

${\log}_2(x^5)=5{\log}_2x$

Zoezi$\PageIndex{3}$

Panua$\ln x^2$.

Jibu

$2\ln x$

Mfano$\PageIndex{4}$: Rewriting an Expression as a Power before Using the Power Rule

Panua${\log}_3(25)$ kutumia utawala wa nguvu kwa magogo.

Suluhisho

Akielezea hoja kama nguvu, tunapata${\log}_3(25)={\log}_3(5^2)$.

Next sisi kutambua exponent$2$,, na msingi,$5$, na kuandika kujieleza sawa na kuzidisha mara exponent logarithm ya msingi.

${\log}_3(52)=2{\log}_3(5)$

Zoezi$\PageIndex{4}$

Panua$\ln\left (\dfrac{1}{x^2} \right )$.

Jibu

$−2\ln(x)$

Mfano$\PageIndex{5}$: Using the Power Rule in Reverse

Andika upya$4\ln(x)$ kwa kutumia utawala wa nguvu kwa magogo kwa logarithm moja na mgawo wa kuongoza wa$1$.

Suluhisho

Kwa sababu logarithm ya nguvu ni bidhaa ya mara exponent logarithm ya msingi, inafuata kwamba bidhaa ya idadi na logarithm inaweza kuandikwa kama nguvu. Kwa kujieleza$4\ln(x)$, sisi kutambua sababu$4$, kama exponent na hoja,$x$, kama msingi, na kuandika upya bidhaa kama logarithm ya nguvu:$4\ln(x)=\ln(x^4)$.

Zoezi$\PageIndex{5}$

Andika upya$2{\log}_34$ kwa kutumia utawala wa nguvu kwa magogo kwa logarithm moja na mgawo wa kuongoza wa$1$.

Jibu

${\log}_316$

Kupanua Maneno ya Logarithmic

Kuchukuliwa pamoja, utawala wa bidhaa, utawala wa quotient, na utawala wa nguvu mara nyingi huitwa “sheria za magogo.” Wakati mwingine tunatumia kanuni zaidi ya moja ili kurahisisha kujieleza. Kwa mfano:

$$\begin{align*} {\log}_b \left (\dfrac{6x}{y} \right )&= {\log}_b(6x)-{\log}_by\\[4pt] &= {\log}_b6+{\log}_bx-{\log}_by \end{align*}$$

Tunaweza kutumia utawala nguvu kupanua maneno logarithmic kuwashirikisha exponents hasi na sehemu. Hapa ni ushahidi mbadala wa utawala wa quotient kwa logarithms kwa kutumia ukweli kwamba usawa ni nguvu hasi:

$$\begin{align*} {\log}_b\left (\dfrac{A}{C} \right )&= {\log}_b(AC^{-1})\\[4pt] &= {\log}_b(A)+{\log}_b(C^{-1})\\[4pt] &= {\log}_bA+(-1){\log}_bC\\[4pt] &= {\log}_bA−{\log}_bC \end{align*}$$

Tunaweza pia kutumia utawala wa bidhaa ili kueleza jumla au tofauti ya logarithms kama logarithm ya bidhaa.

Kwa mazoezi, tunaweza kuangalia kujieleza kwa logarithmic na kupanua kiakili, kuandika jibu la mwisho. Kumbuka, hata hivyo, kwamba tunaweza tu kufanya hivyo kwa bidhaa, quotients, nguvu, na mizizi-kamwe na kuongeza au kutoa ndani ya hoja ya logarithm.

Mfano$\PageIndex{6}$: Expanding Logarithms Using Product, Quotient, and Power Rules

Andika upya$ln \left (\dfrac{x^4y}{7} \right )$ kama jumla au tofauti ya magogo.

Suluhisho

Kwanza, kwa sababu tuna quotient ya maneno mawili, tunaweza kutumia utawala wa quotient:

$\ln \left (\dfrac{x^4y}{7} \right )=\ln(x^4y)−\ln(7)$

Kisha kuona bidhaa katika muda wa kwanza, tunatumia utawala wa bidhaa:

$\ln(x^4y)−\ln(7)=\ln(x^4)+\ln(y)−\ln(7)$

Hatimaye, tunatumia utawala wa nguvu kwa muda wa kwanza:

$\ln(x^4)+\ln(y)−\ln(7)=4\ln(x)+\ln(y)−\ln(7)$

Zoezi$\PageIndex{6}$

Panua$\log \left (\dfrac{x^2y^3}{z^4} \right )$.

Jibu

$2\log x+3\log y−4\log z$

Mfano$\PageIndex{7}$: Using the Power Rule for Logarithms to Simplify the Logarithm of a Radical Expression

Panua$\log(x)$.

Suluhisho

$$\begin{align*} \log(\sqrt{x})&= \log x^{\left (\tfrac{1}{2} \right )}\\[4pt] &= \dfrac{1}{2}\log x \end{align*}$$

Zoezi$\PageIndex{7}$

Panua$\ln(\sqrt[3]{x^2})$.

Jibu

$\dfrac{2}{3}\ln x$

Q & A: Je, tunaweza kupanua$\ln(x^2+y^2)$?

Hapana. Hakuna njia ya kupanua logarithm ya jumla au tofauti ndani ya hoja ya logarithm.

Mfano$\PageIndex{8}$: Expanding Complex Logarithmic Expressions

Panua${\log}_6 \left (\dfrac{64x^3(4x+1)}{(2x−1)} \right )$.

Suluhisho

Tunaweza kupanua kwa kutumia Bidhaa na Quotient Kanuni.

$$\begin{align*} {\log}_6\left (\dfrac{64x^3(4x+1)}{(2x-1)} \right )&= {\log}_664+{\log}_6x^3+{\log}_6(4x+1)-{\log}_6(2x-1)\qquad \text{Apply the Quotient Rule}\\[4pt] &= {\log}_626+{\log}_6x^3+{\log}_6(4x+1)-{\log}_6(2x-1) \qquad \text{Simplify by writing 64 as } 2^6\\[4pt] &= 6{\log}_62+3{\log}_6x+{\log}_6(4x+1)-{\log}_6(2x-1) \qquad \text{Apply the Power Rule} \end{align*}$$

Zoezi$\PageIndex{8}$

Panua$\ln \left (\dfrac{\sqrt{(x−1){(2x+1)}^2}}{(x^2−9)}\right )$.

Jibu

$\dfrac{1}{2}\ln(x−1)+\ln(2x+1)−\ln(x+3)−\ln(x−3)$

Kukondosha Maneno ya Logarithmic

Tunaweza kutumia sheria za logarithms ambazo tumejifunza tu kuzidisha kiasi, tofauti, na bidhaa zilizo na msingi sawa na logarithm moja. Ni muhimu kukumbuka kwamba logarithms lazima iwe na msingi sawa wa kuunganishwa. Tutajifunza baadaye jinsi ya kubadilisha msingi wa logarithm yoyote kabla ya kufuta.

Jinsi ya: Kutokana na jumla, tofauti, au bidhaa ya logarithms na msingi sawa, kuandika kujieleza sawa kama logarithm moja

1. Tumia mali ya nguvu kwanza. Tambua maneno ambayo ni bidhaa za mambo na logarithm, na uandike upya kila mmoja kama logarithm ya nguvu.
2. Ifuatayo tumia mali ya bidhaa. Andika upya kiasi cha logarithms kama logarithm ya bidhaa.
3. Tumia mali ya quotient mwisho. Andika upya tofauti za logarithms kama logarithm ya quotient.

Mfano$\PageIndex{9}$: Using the Product and Quotient Rules to Combine Logarithms

Andika${\log}_3(5)+{\log}_3(8)−{\log}_3(2)$ kama logarithm moja.

Suluhisho

Kutumia bidhaa na sheria za quotient

${\log}_3(5)+{\log}_3(8)={\log}_3(5⋅8)={\log}_3(40)$

Hii inapunguza kujieleza yetu ya awali kwa

${\log}_3(40)−{\log}_3(2)$

Kisha, kwa kutumia utawala wa quotient

${\log}_3(40)−{\log}_3(2)={\log}_3 \left (\dfrac{40}{2} \right )={\log}_3(20)$

Zoezi$\PageIndex{9}$

Condense${\log}_3−{\log}_4+{\log}_5−{\log}_6$.

Jibu

$\log \left (\dfrac{3⋅5}{4⋅6} \right)$; inaweza pia kuandikwa$\log \left (\dfrac{5}{8} \right )$ kwa kupunguza sehemu kwa maneno ya chini kabisa.

Mfano$\PageIndex{10}$: Condensing Complex Logarithmic Expressions

Condense${\log}_2(x^2)+\dfrac{1}{2}{\log}_2(x−1)−3{\log}_2({(x+3)}^2)$.

Suluhisho

Tunatumia utawala wa nguvu kwanza:

${\log}_2(x^2)+\dfrac{1}{2}{\log}_2(x−1)−3{\log}_2({(x+3)}^2)={\log}_2(x^2)+{\log}_2(\sqrt{x−1})−{\log}_2({(x+3)}^6)$

Halafu tunatumia utawala wa bidhaa kwa jumla:

${\log}_2(x^2)+{\log}_2(\sqrt{x−1})−{\log}_2({(x+3)}^6)={\log}_2(x^2\sqrt{x−1})−{\log}_2({(x+3)}^6)$

Hatimaye, tunatumia utawala wa quotient kwa tofauti:

${\log}_2(x^2\sqrt{x−1})−{\log}_2({(x+3)}^6)={\log}_2\dfrac{x^2\sqrt{x−1}}{{(x+3)}^6}$

Mfano$\PageIndex{11}$: Rewriting as a Single Logarithm

Andika upya$2\log x−4\log(x+5)+\dfrac{1}{x}\log(3x+5)$ kama logarithm moja.

Suluhisho

Tunatumia utawala wa nguvu kwanza:

$2\log x−4\log(x+5)+\dfrac{1}{x}\log(3x+5)=\log(x^2)−\log{(x+5)}^4+\log({(3x+5)}^{x^{−1}})$

Halafu tunapanga upya na kutumia utawala wa bidhaa kwa jumla:

$$\begin{align*} \log(x^2)-\log{(x+5)}^4+\log({(3x+5)}^{x^{-1}})&= \log(x^2)+\log({(3x+5)}^{x^{-1}}-\log{(x+5)}^4\\[4pt] &= \log(x^2{(3x+5)}^{x^{-1}})-\log{(x+5)}^4\\[4pt] &= \log\dfrac{x^2{(3x+5)}^{x^{-1}}}{{(x+5)}^4} \qquad \text{Apply the quotient rule to the difference} \end{align*}$$

Zoezi$\PageIndex{10}$

Andika upya$\log(5)+0.5\log(x)−\log(7x−1)+3\log(x−1)$ kama logarithm moja.

Jibu

$\log \dfrac{5{(x−1)}^3\sqrt{x}}{(7x−1)}$

Zoezi$\PageIndex{11}$

Condense$4(3\log(x)+\log(x+5)−\log(2x+3))$.

Jibu

$\log\dfrac{x^{12}{(x+5)}^4}{{(2x+3)}^4}$; jibu hili linaweza pia kuandikwa$\log{ \left (\dfrac{x^3(x+5)}{(2x+3)} \right )}^4$

Mfano$\PageIndex{12}$: Applying of the Laws of Logs

Kumbuka kwamba, katika kemia,${pH}=−\log[H+]$. Ikiwa mkusanyiko wa ions hidrojeni katika kioevu ni mara mbili, ni athari gani kwenye pH?

Suluhisho

Tuseme$C$ ni mkusanyiko wa awali wa ions hidrojeni, na$P$ ni pH ya awali ya kioevu. Kisha$P=–\log(C)$. Ikiwa mkusanyiko umeongezeka mara mbili, ukolezi mpya ni$2C$. Kisha pH ya kioevu mpya ni

$pH=−\log(2C)$

Kutumia utawala wa bidhaa wa magogo

$pH=−\log(2C)=−(\log(2)+\log(C))=−\log(2)−\log(C)$

Tangu$P=–\log(C)$, pH mpya ni

$pH=P−\log(2)≈P−0.301$

Zoezi$\PageIndex{12}$

Wakati mkusanyiko wa ions hidrojeni ni mara mbili, pH inapungua kwa karibu$0.301$.

Je, pH inabadilikaje wakati mkusanyiko wa ions hidrojeni chanya unapungua kwa nusu?

Jibu

PH huongezeka kwa karibu$0.301$.

Kutumia Mfumo wa Mabadiliko-ya-msingi kwa Logarithms

Wahesabu wengi wanaweza kutathmini magogo ya kawaida na ya asili. Ili kutathmini logarithms na msingi zaidi ya$10$ ore, e, tunatumia formula ya mabadiliko-ya msingi ili kuandika upya logarithm kama quotient ya logarithms ya msingi mwingine wowote; wakati wa kutumia calculator, tunataka kubadili yao kwa magogo ya kawaida au ya asili.

Ili kupata formula ya mabadiliko-msingi, tunatumia mali moja kwa moja na utawala wa nguvu kwa logarithms.

Kutokana na idadi yoyote chanya halisi$M$$b$,, na$n$, wapi$n≠1$ na$b≠1$, sisi kuonyesha

${\log}_bM=\dfrac{{\log}_nM}{{\log}_nb}$

Hebu$y={\log}_bM$. Kwa kuchukua msingi wa logi$n$ wa pande zote mbili za equation, tunakuja fomu ya kielelezo, yaani$b^y=M$. Inafuata kwamba

$$\begin{align*} {\log}_n(b^y)&= {\log}_nM \qquad \text{Apply the one-to-one property}\\[4pt] y{\log}_nb&= {\log}_nM \qquad \text{Apply the power rule for logarithms}\\[4pt] y&= \dfrac{{\log}_nM}{{\log}_nb} \qquad \text{Isolate y}\\[4pt] {\log}_bM&= \dfrac{{\log}_nM}{{\log}_nb} \qquad \text{Substitute for y} \end{align*}$$

Kwa mfano, kutathmini${\log}_536$ kutumia calculator, lazima kwanza kuandika upya maneno kama quotient ya magogo ya kawaida au ya asili. Tutatumia logi ya kawaida.

$$\begin{align*} {\log}_536&= \dfrac{\log(36)}{\log(5)} \qquad \text{Apply the change of base formula using base 10}\\[4pt] &\approx 2.2266 \qquad \text{Use a calculator to evaluate to 4 decimal places} \end{align*}$$

FOMU YA MABADILIKO YA MSINGI

Fomu ya mabadiliko-msingi inaweza kutumika kutathmini logarithm na msingi wowote.

Kwa idadi yoyote chanya halisi$M$,$b$, na$n$, wapi$n≠1$ na$b≠1$,

$${\log}_bM=\dfrac{{\log}_nM}{{\log}_nb}$$

Inafuata kwamba formula ya mabadiliko-ya msingi inaweza kutumika kuandika upya logarithm na msingi wowote kama quotient ya magogo ya kawaida au ya asili.

$${\log}_bM=\dfrac{\ln M}{\ln b}$$

na

$${\log}_bM=\dfrac{\log M}{\log b}$$

Jinsi ya: Kutokana na logarithm na fomu${\log}_bM$, use the change-of-base formula to rewrite it as a quotient of logs with any positive base $n$, where $n≠1$

1. Kuamua msingi mpya$n$, kukumbuka kuwa logi ya kawaida$\log(x)$, ina msingi wa 10, na logi ya asili$\ln(x)$, ina msingi$e$.
2. Andika upya logi kama quotient kutumia formula ya mabadiliko-msingi
   • Nambari ya quotient itakuwa logarithm na msingi$n$ na hoja$M$.
   • Denominator ya quotient itakuwa logarithm na msingi$n$ na hoja$b$.

Mfano$\PageIndex{13}$: Changing Logarithmic Expressions to Expressions Involving Only Natural Logs

Badilisha${\log}_53$ kwa quotient ya logarithms asili.

Suluhisho

Kwa sababu tutakuwa kuonyesha${\log}_53$ kama quotient ya logarithms asili, msingi mpya,$n=e$.

Tunaandika upya logi kama quotient kutumia formula ya mabadiliko-msingi. Nambari ya quotient itakuwa logi ya asili na hoja$3$. Denominator ya quotient itakuwa logi ya asili na hoja 5.

${\log}_bM=\dfrac{\ln M}{\ln b}$

${\log}_53=\dfrac{\ln3}{\ln5}$

Zoezi$\PageIndex{13}$

Badilisha$\log0.58$ kwa quotient ya logarithms asili.

Jibu

$\dfrac{\ln8}{\ln 0.5}$

Q & A: Je, tunaweza kubadilisha logarithms kawaida kwa logarithms asili?

Ndiyo. Kumbuka kwamba$\log9$ maana${\log}_{10}9$. Hivyo,$\log9=\dfrac{\ln9}{\ln10}$.

Mfano$\PageIndex{14}$: Using the Change-of-Base Formula with a Calculator

Tathmini${\log}_2(10)$ kutumia formula ya mabadiliko-ya msingi na calculator.

Suluhisho

Kwa mujibu wa formula ya mabadiliko-msingi, tunaweza kuandika upya msingi wa logi$2$ kama logarithm ya msingi mwingine wowote. Kwa kuwa mahesabu yetu yanaweza kutathmini logi ya asili, tunaweza kuchagua kutumia logarithm ya asili, ambayo ni msingi wa logi$e$.

$$\begin{align*} {\log}_210&= \dfrac{\ln10}{\ln2} \qquad \text{Apply the change of base formula using base } e\\[4pt] &\approx 3.3219 \qquad \text{Use a calculator to evaluate to 4 decimal places} \end{align*}$$

Zoezi$\PageIndex{14}$

Tathmini${\log}_5(100)$ kutumia formula ya mabadiliko-ya msingi.

Jibu

$\dfrac{\ln100}{\ln5}≈\dfrac{4.6051}{1.6094}=2.861$

Media

Fikia rasilimali hizi mtandaoni kwa maelekezo ya ziada na ufanyie mazoezi na sheria za logarithms.

• Mali ya Logarithms
• Panua Maneno ya Logarithmic
• Tathmini ya kujieleza ya asili ya Logarithmic

Mlinganyo muhimu

Utawala wa Bidhaa kwa Logarithms	${\log}_b(MN)={\log}_b(M)+{\log}_b(N)$
Utawala wa Quotient kwa Logarithms	${\log}_b(\dfrac{M}{N})={\log}_bM−{\log}_bN$
Utawala wa Nguvu kwa Logarithms	${\log}_b(M^n)=n{\log}_bM$
Mfumo wa Mabadiliko-ya-msingi	${\log}_bM=\dfrac{{\log}_nM}{{\log}_nb}$$n>0$,$n≠1$,$b≠1$

Dhana muhimu

• Tunaweza kutumia utawala wa bidhaa wa logarithms kuandika upya logi ya bidhaa kama jumla ya logarithms. Angalia Mfano$\PageIndex{1}$.
• Tunaweza kutumia utawala wa quotient wa logarithms kuandika upya logi ya quotient kama tofauti ya logarithms. Angalia Mfano$\PageIndex{2}$.
• Tunaweza kutumia utawala wa nguvu kwa logarithms kuandika upya logi ya nguvu kama bidhaa ya exponent na logi ya msingi wake. Angalia Mfano$\PageIndex{3}$, Mfano$\PageIndex{4}$, na Mfano$\PageIndex{5}$.
• Tunaweza kutumia utawala wa bidhaa, utawala wa quotient, na utawala wa nguvu pamoja ili kuchanganya au kupanua logarithm na pembejeo tata. Angalia Mfano$\PageIndex{6}$, Mfano$\PageIndex{7}$, na Mfano$\PageIndex{8}$.
• Sheria za logarithms pia zinaweza kutumiwa kupunguza kiasi, tofauti, na bidhaa zilizo na msingi sawa na logarithm moja. Angalia Mfano$\PageIndex{9}$, Mfano$\PageIndex{10}$, Mfano$\PageIndex{11}$, na Mfano$\PageIndex{12}$.
• Tunaweza kubadilisha logarithm na msingi wowote kwa quotient ya logarithms na msingi nyingine yoyote kwa kutumia mabadiliko ya-msingi formula. Angalia Mfano$\PageIndex{13}$.
• Fomu ya mabadiliko-msingi mara nyingi hutumiwa kuandika upya logarithm na msingi zaidi ya 10 na$e$ kama quotient ya magogo ya asili au ya kawaida. Kwa njia hiyo calculator inaweza kutumika kutathmini. Angalia Mfano$\PageIndex{14}$.