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10.5: Tumia Mali ya Logarithms

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    Malengo ya kujifunza

    Mwishoni mwa sehemu hii, utaweza:

    • Tumia mali ya logarithms
    • Tumia Mabadiliko ya Mfumo wa Msingi

    Kabla ya kuanza, fanya jaribio hili la utayari.

    1. Tathmini: a.\(a^{0}\) b\(a^{1}\).
      Kama amekosa tatizo hili, mapitio Mfano 5.14.
    2. Andika kwa kielelezo cha busara:\(\sqrt[3]{x^{2} y}\).
      Kama amekosa tatizo hili, mapitio Mfano 8.27.
    3. Pande zote hadi maeneo matatu ya decimal:\(2.5646415\).
      Kama amekosa tatizo hili, mapitio Mfano 1.34.

    Tumia Mali ya Logarithms

    Sasa kwa kuwa tumejifunza kuhusu kazi za kielelezo na za logarithmic, tunaweza kuanzisha baadhi ya mali za logarithms. Hizi zitasaidia sana kama tunaendelea kutatua equations wote kielelezo na logarithmic.

    Mali mbili za kwanza zinatokana na ufafanuzi wa logarithms. Tangu\(a^{0}=1\), tunaweza kubadilisha hii kwa fomu ya logarithmic na kupata\(\log _{a} 1=0\). Pia, tangu\(a^{1}=a\), tunapata\(\log _{a} a=1\).

    Ufafanuzi\(\PageIndex{1}\)

    Mali ya Logarithms

    \(\log _{a} 1=0 \quad \log _{a} a=1\)

    Katika mfano unaofuata tunaweza kutathmini logarithm kwa kuwabadili fomu ya kielelezo, kama tulivyofanya hapo awali, lakini kutambua na kisha kutumia mali anaokoa muda.

    Mfano\(\PageIndex{1}\)

    Tathmini kutumia mali ya logarithms:

    1. \(\log _{8} 1\)
    2. \(\log _{6} 6\)

    Suluhisho:

    a.

    \(\log _{8} 1\)

    Tumia mali,\(\log _{a} 1=0\).

    \(0 \quad \log _{8} 1=0\)

    b.

    \(\log _{6} 6\)

    Tumia mali,\(\log _{a} a=1\).

    \(1 \quad \log _{6} 6=1\)

    Zoezi\(\PageIndex{1}\)

    Tathmini kutumia mali ya logarithms:

    1. \(\log _{13} 1\)
    2. \(\log _{9} 9\)
    Jibu
    1. \(0\)
    2. \(1\)
    Zoezi\(\PageIndex{2}\)

    Tathmini kutumia mali ya logarithms:

    1. \(\log _{5} 1\)
    2. \(\log _{7} 7\)
    Jibu
    1. \(0\)
    2. \(1\)

    Mali mbili zifuatazo zinaweza kuthibitishwa kwa kuwabadili kutoka fomu ya kielelezo hadi fomu ya logarithmic, au kinyume.

    Equation ya kielelezo\(a^{\log _{a} x}=x\) inabadilisha equation ya logarithmic\(\log _{a} x=\log _{a} x\), ambayo ni taarifa ya kweli kwa maadili mazuri kwa\(x\) tu.

    Equation ya logarithmic\(\log _{a} a^{x}=x\) inabadilisha equation ya kielelezo\(a^{x}=a^{x}\), ambayo pia ni taarifa ya kweli.

    Mali hizi mbili huitwa mali inverse kwa sababu, wakati tuna msingi huo, kuinua kwa nguvu “kufuta” logi na kuchukua logi “kufuta” kuinua kwa nguvu. Mali hizi mbili zinaonyesha muundo wa kazi. Wote kuishia na kazi ya utambulisho ambayo inaonyesha tena kwamba kazi kielelezo na logarithmic ni kazi inverse.

    Ufafanuzi\(\PageIndex{2}\)

    Mali ya Inverse ya Logarithms

    Kwa\(a>0, x>0\) na\(a \neq 1\),

    \(a^{\log _{a} x}=x \quad \log _{a} a^{x}=x\)

    Katika mfano unaofuata, tumia mali ya inverse ya logarithms.

    Mfano\(\PageIndex{2}\)

    Tathmini kutumia mali ya logarithms:

    1. \(4^{\log _{4} 9}\)
    2. \(\log _{3} 3^{5}\)

    Suluhisho:

    a.

    \(4^{\log _{4} 9}\)

    Tumia mali,\(a^{\log _{a} x}=x\).

    \(9 \quad 4^{\log _{4} 9}=9\)

    b.

    \(\log _{3} 3^{5}\)

    Tumia mali,\(a^{\log _{a} x}=x\).

    \(5 \quad \log _{3} 3^{5}=5\)

    Zoezi\(\PageIndex{3}\)

    Tathmini kutumia mali ya logarithms:

    1. \(5^{\log _{5} 15}\)
    2. \(\log _{7} 7^{4}\)
    Jibu
    1. \(15\)
    2. \(4\)
    Zoezi\(\PageIndex{4}\)

    Tathmini kutumia mali ya logarithms:

    1. \(2^{\log _{2} 8}\)
    2. \(\log _{2} 2^{15}\)
    Jibu
    1. \(8\)
    2. \(15\)

    Kuna mali tatu zaidi za logarithms ambazo zitafaa katika kazi yetu. Tunajua kazi za kielelezo na kazi ya logarithmic zinahusiana sana. Ufafanuzi wetu wa logarithm unatuonyesha kwamba logarithm ni kielelezo cha kielelezo sawa. mali ya exponents na mali kuhusiana kwa exponents.

    Katika Bidhaa Mali ya Exponents\(a^{m} \cdot a^{n}=a^{m+n}\), tunaona kwamba kuzidisha msingi huo, sisi kuongeza exponents. Mali ya Bidhaa ya Logarithms,\(\log _{a} M \cdot N=\log _{a} M+\log _{a} N\) inatuambia kuchukua logi ya bidhaa, tunaongeza logi ya mambo.

    Ufafanuzi\(\PageIndex{3}\)

    Bidhaa Mali ya Logarithms

    Kama\(M>0, N>0, \mathrm{a}>0\) na\(\mathrm{a} \neq 1,\) kisha

    \(\log _{a}(M \cdot N)=\log _{a} M+\log _{a} N\)

    Logarithm ya bidhaa ni jumla ya logarithms.

    Tunatumia mali hii kuandika logi ya bidhaa kama jumla ya magogo ya kila sababu.

    Mfano\(\PageIndex{3}\)

    Tumia Mali ya Bidhaa ya Logarithms kuandika kila logarithm kama jumla ya logarithms. Kurahisisha, ikiwa inawezekana:

    1. \(\log _{3} 7 x\)
    2. \(\log _{4} 64 x y\)

    Suluhisho:

    a.

    \(\log _{3} 7 x\)

    Tumia Mali ya Bidhaa,\(\log _{a}(M \cdot N)=\log _{a} M+\log _{a} N\).

    \(\log _{3} 7+\log _{3} x\)
    \(\log _{3} 7 x=\log _{3} 7+\log _{3} x\)

    b.

    \(\log _{4} 64 x y\)

    Tumia Mali ya Bidhaa,\(\log _{a}(M \cdot N)=\log _{a} M+\log _{a} N\).

    \(\log _{4} 64+\log _{4} x+\log _{4} y\)

    Kurahisisha kuwa kutathmini,\(\log _{4} 64\).

    \(3+\log _{4} x+\log _{4} y\)
    \(\log _{4} 64 x y=3+\log _{4} x+\log _{4} y\)

    Zoezi\(\PageIndex{5}\)

    Tumia Mali ya Bidhaa ya Logarithms kuandika kila logarithm kama jumla ya logarithms. Kurahisisha, ikiwa inawezekana:

    1. \(\log _{3} 3 x\)
    2. \(\log _{2} 8 x y\)
    Jibu
    1. \(1+\log _{3} x\)
    2. \(3+\log _{2} x+\log _{2} y\)
    Zoezi\(\PageIndex{6}\)

    Tumia Mali ya Bidhaa ya Logarithms kuandika kila logarithm kama jumla ya logarithms. Kurahisisha, ikiwa inawezekana:

    1. \(\log _{9} 9 x\)
    2. \(\log _{3} 27 x y\)
    Jibu
    1. \(1+\log _{9} x\)
    2. \(3+\log _{3} x+\log _{3} y\)

    Vile vile, katika Quotient Mali ya Exponents\(\frac{a^{m}}{a^{n}}=a^{m-n}\), tunaona kwamba kugawanya msingi huo, sisi Ondoa exponents. Mali ya Quotient ya Logarithms,\(\log _{a} \frac{M}{N}=\log _{a} M-\log _{a} N\) inatuambia kuchukua logi ya quotient, tunaondoa logi ya nambari na denominator.

    Ufafanuzi\(\PageIndex{4}\)

    Mali ya Quotient ya Logarithms

    Kama\(M>0, N>0, \mathrm{a}>0\) na\(\mathrm{a} \neq 1,\) kisha

    \(\log _{a} \frac{M}{N}=\log _{a} M-\log _{a} N\)

    Logarithm ya quotient ni tofauti ya logarithms.

    Kumbuka kwamba\(\log _{a} M=\log _{a} N \not=\log _{a}(M-N)\).

    Tunatumia mali hii kuandika logi ya quotient kama tofauti ya magogo ya kila sababu.

    Mfano\(\PageIndex{4}\)

    Tumia Mali ya Quotient ya Logarithms kuandika kila logarithm kama tofauti ya logarithms. Kurahisisha, ikiwa inawezekana.

    1. \(\log _{5} \frac{5}{7}\)
    2. \(\log \frac{x}{100}\)

    Suluhisho:

    a.

    \(\log _{5} \frac{5}{7}\)

    Tumia Mali ya Quotient,\(\log _{a} \frac{M}{N}=\log _{a} M-\log _{a} N\).

    \(\log _{5} 5-\log _{5} 7\)

    Kurahisisha.

    \(1-\log _{5} 7\)

    \(\log _{5} \frac{5}{7}=1-\log _{5} 7\)

    b.

    \(\log \frac{x}{100}\)

    Tumia Mali ya Quotient,\(\log _{a} \frac{M}{N}=\log _{a} M-\log _{a} N\).

    \(\log x-\log 100\)

    Kurahisisha.

    \(\log x-2\)

    \(\log \frac{x}{100}=\log x-2\)

    Zoezi\(\PageIndex{7}\)

    Tumia Mali ya Quotient ya Logarithms kuandika kila logarithm kama tofauti ya logarithms. Kurahisisha, ikiwa inawezekana.

    1. \(\log _{4} \frac{3}{4}\)
    2. \(\log \frac{x}{1000}\)
    Jibu
    1. \(\log _{4} 3-1\)
    2. \(\log x-3\)
    Zoezi\(\PageIndex{8}\)

    Tumia Mali ya Quotient ya Logarithms kuandika kila logarithm kama tofauti ya logarithms. Kurahisisha, ikiwa inawezekana.

    1. \(\log _{2} \frac{5}{4}\)
    2. \(\log \frac{10}{y}\)
    Jibu
    1. \(\log _{2} 5-2\)
    2. \(1-\log y\)

    mali ya tatu ya logarithms ni kuhusiana na Power Mali ya Exponents\(\left(a^{m}\right)^{n}=a^{m \cdot n}\), tunaona kwamba kuongeza nguvu kwa nguvu, sisi kuzidisha exponents. Mali ya Nguvu ya Logarithms,\(\log _{a} M^{p}=p \log _{a} M\) inatuambia kuchukua logi ya nambari iliyoinuliwa kwa nguvu, tunazidisha mara za nguvu logi ya nambari.

    Ufafanuzi\(\PageIndex{5}\)

    Nguvu Mali ya Logarithms

    Kama\(M>0, \mathrm{a}>0, \mathrm{a} \neq 1\) na\(p\) ni idadi yoyote halisi basi,

    \(\log _{a} M^{p}=p \log _{a} M\)

    Logi ya nambari iliyoinuliwa kwa nguvu kama bidhaa za bidhaa za nguvu mara logi ya nambari.

    Tunatumia mali hii kuandika logi ya nambari iliyoinuliwa kwa nguvu kama bidhaa ya nguvu mara logi ya nambari. Sisi kimsingi kuchukua exponent na kutupa mbele ya logarithm.

    Mfano\(\PageIndex{5}\)

    Tumia Mali ya Nguvu ya Logarithms kuandika kila logarithm kama bidhaa ya logarithms. Kurahisisha, ikiwa inawezekana.

    1. \(\log _{5} 4^{3}\)
    2. \(\log x^{10}\)

    Suluhisho:

    a.

    \(\log _{5} 4^{3}\)

    Tumia Mali ya Nguvu,\(\log _{a} M^{p}=p \log _{a} M\).

    3\(\log _{5} 4\)

    \(\log _{5} 4^{3}=3 \log _{5} 4\)

    b.

    \(\log x^{10}\)

    Tumia Mali ya Nguvu,\(\log _{a} M^{p}=p \log _{a} M\).

    \(10\log x\)

    \(\log x^{10}=10 \log x\)

    Zoezi\(\PageIndex{9}\)

    Tumia Mali ya Nguvu ya Logarithms kuandika kila logarithm kama bidhaa ya logarithms. Kurahisisha, ikiwa inawezekana.

    1. \(\log _{7} 5^{4}\)
    2. \(\log x^{100}\)
    Jibu
    1. \(4\log _{7} 5\)
    2. 100\(\cdot \log x\)
    Zoezi\(\PageIndex{10}\)

    Tumia Mali ya Nguvu ya Logarithms kuandika kila logarithm kama bidhaa ya logarithms. Kurahisisha, ikiwa inawezekana.

    1. \(\log _{2} 3^{7}\)
    2. \(\log x^{20}\)
    Jibu
    1. \(7\log _{2} 3\)
    2. \(20\cdot \log x\)

    Sisi muhtasari Mali ya Logarithms hapa kwa ajili ya kumbukumbu rahisi. Wakati logarithms asili ni kesi maalum ya mali hizi, mara nyingi ni muhimu pia kuonyesha asili logarithm version ya kila mali.

    Mali ya Logarithms

    Kama\(M>0, \mathrm{a}>0, \mathrm{a} \neq 1\) na\(p\) ni idadi yoyote halisi basi,

    Mali Msingi\(a\) Msingi\(e\)
    \ (a\) ">\(\log _{a} 1=0\) \ (e\) ">\(\ln 1=0\)
    \ (a\) ">\(\log _{a} a=1\) \ (e\) ">\(\ln e=1\)
    Mali Inverse \ (a\) ">\(a^{\log _{a} x}=x\)
    \(\log _{a} a^{x}=x\)    		 \ (e\) ">\(e^{\ln x}=x\)
    \(\ln e^{x}=x\)
    Bidhaa Mali ya Logarithms \ (a\) ">\(\log _{a}(M \cdot N)=\log _{a} M+\log _{a} N\) \ (e\) ">\(\ln (M \cdot N)=\ln M+\ln N\)
    Mali ya Quotient ya Logarithms \ (a\) ">\(\log _{a} \frac{M}{N}=\log _{a} M-\log _{a} N\) \ (e\) ">\(\ln \frac{M}{N}=\ln M-\ln N\)
    Nguvu Mali ya Logarithms \ (a\) ">\(\log _{a} M^{p}=p \log _{a} M\) \ (e\) ">\(\ln M^{p}=p \ln M\)
    Jedwali 10.4.1

    Sasa kwa kuwa tuna mali tunaweza kuzitumia “kupanua” kujieleza kwa logarithmic. Hii inamaanisha kuandika logarithm kama jumla au tofauti na bila mamlaka yoyote.

    Kwa ujumla tunatumia Bidhaa na Quotient Properties kabla ya kutumia Power Property.

    Mfano\(\PageIndex{6}\)

    Tumia Mali ya Logarithms kupanua logarithm\(\log _{4}\left(2 x^{3} y^{2}\right)\). Kurahisisha, ikiwa inawezekana.

    Suluhisho:

    Tumia Mali ya Bidhaa,\(\log _{a} M \cdot N=\log _{a} M+\log _{a} N\).

    Tumia Mali ya Nguvu\(\log _{a} M^{p}=p \log _{a} M\),, kwa maneno mawili ya mwisho. Kurahisisha.

    Zoezi\(\PageIndex{11}\)

    Tumia Mali ya Logarithms kupanua logarithm\(\log _{2}\left(5 x^{4} y^{2}\right)\). Kurahisisha, ikiwa inawezekana.

    Jibu

    \(\log _{2} 5+4 \log _{2} x+2 \log _{2} y\)

    Zoezi\(\PageIndex{12}\)

    Tumia Mali ya Logarithms kupanua logarithm\(\log _{3}\left(7 x^{5} y^{3}\right)\). Kurahisisha, ikiwa inawezekana.

    Jibu

    \(\log _{3} 7+5 \log _{3} x+3 \log _{3} y\)

    Wakati tuna radical katika kujieleza logarithmic, ni muhimu kwanza kuandika radicand yake kama exponent busara.

    Mfano\(\PageIndex{7}\)

    Tumia Mali ya Logarithms kupanua logarithm\(\log _{2} \sqrt[4]{\frac{x^{3}}{3 y^{2} z}}\). Kurahisisha, ikiwa inawezekana.

    Suluhisho

    \(\log _{2} \sqrt[4]{\frac{x^{3}}{3 y^{2} z}}\)

    Andika upya radical na exponent busara.

    \(\log _{2}\left(\frac{x^{3}}{3 y^{2} z}\right)^{\frac{1}{4}}\)

    Tumia Mali ya Nguvu,\(\log _{a} M^{p}=p \log _{a} M\).

    \(\frac{1}{4} \log _{2}\left(\frac{x^{3}}{3 y^{2} z}\right)\)

    Tumia Mali ya Quotient,\(\log _{a} M \cdot N=\log _{a} M-\log _{a} N\).

    \(\frac{1}{4}\left(\log _{2}\left(x^{3}\right)-\log _{2}\left(3 y^{2} z\right)\right)\)

    Tumia Mali ya Bidhaa\(\log _{a} M \cdot N=\log _{a} M+\log _{a} N\),, katika kipindi cha pili.

    \(\frac{1}{4}\left(\log _{2}\left(x^{3}\right)-\left(\log _{2} 3+\log _{2} y^{2}+\log _{2} z\right)\right)\)

    Tumia Mali ya Nguvu\(\log _{a} M^{p}=p \log _{a} M\), ndani ya mabano.

    \(\frac{1}{4}\left(3 \log _{2} x-\left(\log _{2} 3+2 \log _{2} y+\log _{2} z\right)\right)\)

    Kurahisisha kwa kusambaza.

    \(\frac{1}{4}\left(3 \log _{2} x-\log _{2} 3-2 \log _{2} y-\log _{2} z\right)\)

    \(\log _{2} \sqrt[4]{\frac{x^{3}}{3 y^{2} z}}=\frac{1}{4}\left(3 \log _{2} x-\log _{2} 3-2 \log _{2} y-\log _{2} z\right)\)

    Zoezi\(\PageIndex{13}\)

    Tumia Mali ya Logarithms kupanua logarithm\(\log _{4} \sqrt[5]{\frac{x^{4}}{2 y^{3} z^{2}}}\). Kurahisisha, ikiwa inawezekana.

    Jibu

    \(\frac{1}{5}\left(4 \log _{4} x-\frac{1}{2}-3 \log _{4} y-2 \log _{4} z\right)\)

    Zoezi\(\PageIndex{14}\)

    Tumia Mali ya Logarithms kupanua logarithm\(\log _{3} \sqrt[3]{\frac{x^{2}}{5 y z}}\). Kurahisisha, ikiwa inawezekana.

    Jibu

    \(\frac{1}{3}\left(2 \log _{3} x-\log _{3} 5-\log _{3} y-\log _{3} z\right)\)

    Kinyume cha kupanua logarithm ni kufungia jumla au tofauti ya logarithms ambayo ina msingi sawa katika logarithm moja. Sisi tena kutumia mali ya logarithms kutusaidia, lakini kwa reverse.

    Ili kuimarisha maneno ya logarithmic na msingi sawa katika logarithm moja, tunaanza kwa kutumia Mali ya Power ili kupata coefficients ya maneno ya logi kuwa moja na kisha Bidhaa na Quotient Mali kama inahitajika.

    Mfano\(\PageIndex{8}\)

    Tumia Mali ya Logarithms ili kuimarisha logarithm\(\log _{4} 3+\log _{4} x-\log _{4} y\). Kurahisisha, ikiwa inawezekana.

    Suluhisho:

    Maneno ya logi yote yana msingi sawa,\(4\).

    Masharti mawili ya kwanza yanaongezwa, kwa hiyo tunatumia Mali ya Bidhaa,\(\log _{a} M+\log _{a} N=\log _{a} M : N\).

    Tangu magogo yameondolewa, tunatumia Mali ya Quotient,\(\log _{a} M-\log _{a} N=\log _{a} \frac{M}{N}\).

    Zoezi\(\PageIndex{15}\)

    Tumia Mali ya Logarithms ili kuimarisha logarithm\(\log _{2} 5+\log _{2} x-\log _{2} y\). Kurahisisha, ikiwa inawezekana.

    Jibu

    \(\log _{2} \frac{5 x}{y}\)

    Zoezi\(\PageIndex{16}\)

    Tumia Mali ya Logarithms ili kuimarisha logarithm\(\log _{3} 6-\log _{3} x-\log _{3} y\). Kurahisisha, ikiwa inawezekana.

    Jibu

    \(\log _{3} \frac{6}{x y}\)

    Mfano\(\PageIndex{9}\)

    Tumia Mali ya Logarithms ili kuimarisha logarithm\(2 \log _{3} x+4 \log _{3}(x+1)\). Kurahisisha, ikiwa inawezekana.

    Suluhisho:

    Maneno ya logi yana msingi sawa,\(3\).

    \(2 \log _{3} x+4 \log _{3}(x+1)\)

    Tumia Mali ya Nguvu,\(\log _{a} M+\log _{a} N=\log _{a} M \cdot N\).

    \(\log _{3} x^{2}+\log _{3}(x+1)^{4}\)

    Masharti yanaongezwa, kwa hiyo tunatumia Mali ya Bidhaa,\(\log _{a} M+\log _{a} N=\log _{a} M \cdot N\).

    \(\log _{3} x^{2}(x+1)^{4}\)
    \(2 \log _{3} x+4 \log _{3}(x+1)=\log _{3} x^{2}(x+1)^{4}\)

    Zoezi\(\PageIndex{17}\)

    Tumia Mali ya Logarithms ili kuimarisha logarithm\(3 \log _{2} x+2 \log _{2}(x-1)\). Kurahisisha, ikiwa inawezekana.

    Jibu

    \(\log _{2} x^{3}(x-1)^{2}\)

    Zoezi\(\PageIndex{18}\)

    Tumia Mali ya Logarithms ili kuimarisha logarithm\(2 \log x+2 \log (x+1)\). Kurahisisha, ikiwa inawezekana.

    Jibu

    \(\log x^{2}(x+1)^{2}\)

    Tumia Mfumo wa Mabadiliko-ya-Msingi

    Ili kutathmini logarithm na msingi mwingine wowote, tunaweza kutumia Mfumo wa Mabadiliko ya Msingi. Tutaonyesha jinsi hii inatokana.

    \(\begin{array} {l c} {\text{Suppose we want to evaluate} \log_{a}M} & {\log_{a}M} \\ {\text{Let} \:y =\log_{a}M. }&{y=\log_{a}M} \\ {\text{Rewrite the epression in exponential form. }}&{a^{y}=M } \\ {\text{Take the }\:\log_{b} \text{of each side.}}&{\log_{b}a^{y}=\log_{b}M}\\ {\text{Use the Power Property.}}&{y\log_{b}a=\log_{b}M} \\ {\text{Solve for}\:y. }&{y=\frac{\log_{b}M}{\log_{b}a}} \\ {\text{Substiture}\:y=\log_{a}M.}&{\log_{a}M=\frac{\log_{b}M}{\log_{b}a}} \end{array}\)

    Mfumo wa Mabadiliko ya Msingi huanzisha msingi mpya\(b\). Hii inaweza kuwa msingi wowote\(b\) tunataka wapi\(b>0,b≠1\). Kwa sababu mahesabu yetu yana funguo za msingi\(10\) na msingi wa logarithms\(e\), tutaandika upya Mfumo wa Mabadiliko-ya-msingi na msingi mpya kama\(10\) au\(e\).

    Ufafanuzi\(\PageIndex{6}\)

    Fomu ya Mabadiliko ya-msingi

    Kwa misingi yoyote ya logarithmic\(a, b\) na\(M>0\),

    \(\begin{array}{lll}{\log _{a} M=\frac{\log _{b} M}{\log _{b} a}} & {\log _{a} M=\frac{\log M}{\log a}} & {\log _{a} M=\frac{\ln M}{\ln a}} \\ {\text { new base } b} & {\text { new base } 10} & {\text { new base } e}\end{array}\)

    Tunapotumia calculator kupata thamani ya logarithm, sisi kawaida pande zote kwa sehemu tatu decimal. Hii inatupa thamani ya takriban na hivyo tunatumia ishara takriban sawa\((≈)\).

    Mfano\(\PageIndex{10}\)

    Kuzunguka kwa maeneo matatu ya decimal, takriban\(\log _{4} 35\).

    Suluhisho:

      .
    Tumia Mfumo wa Mabadiliko-ya-msingi. .
    Tambua\(a\) na\(M\). Chagua\(10\) kwa ajili ya\(b\). .
    Ingiza maneno\(\frac{\log 35}{\log 4}\) katika calculator kutumia kifungo logi kwa msingi\(10\). Pande zote hadi maeneo matatu ya decimal. .
    Jedwali 10.4.2
    Zoezi\(\PageIndex{19}\)

    Kuzunguka kwa maeneo matatu ya decimal, takriban\(\log _{3} 42\).

    Jibu

    \(3.402\)

    Zoezi\(\PageIndex{20}\)

    Kuzunguka kwa maeneo matatu ya decimal, takriban\(\log _{5} 46\).

    Jibu

    \(2.379\)

    Fikia rasilimali hizi za mtandaoni kwa maelekezo ya ziada na mazoezi kwa kutumia mali ya logarithms.

    Dhana muhimu

    • \(\log _{a} 1=0 \quad \log _{a} a=1\)
    • Mali ya Inverse ya Logarithms
      • Kwa\(a>0,x>0\) na\(a≠1\)

        \(a^{\log _{a} x}=x \quad \log _{a} a^{x}=x\)

    • Bidhaa Mali ya Logarithms
      • Ikiwa\(M>0,N>0,a>0\) na\(a≠1\), basi,

        \(\log _{a} M \cdot N=\log _{a} M+\log _{a} N\)

        Logarithm ya bidhaa ni jumla ya logarithms.
    • Mali ya Quotient ya Logarithms
      • Ikiwa\(M>0, N>0, \mathrm{a}>0\) na\(a≠1\), basi,

        \(\log _{a} \frac{M}{N}=\log _{a} M-\log _{a} N\)

        Logarithm ya quotient ni tofauti ya logarithms.
    • Nguvu Mali ya Logarithms
      • Kama\(M>0,a>0,a≠1\) na\(p\) ni idadi yoyote halisi basi,

        \(\log _{a} M^{p}=p \log _{a} M\)

        Logi ya nambari iliyoinuliwa kwa nguvu ni bidhaa ya nguvu mara logi ya nambari.
    • Mali ya Logarithms Summary
      Kama\(M>0,a>0,a≠1\) na\(p\) ni idadi yoyote halisi basi,
    Mali Msingi\(a\) Msingi\(e\)
    \ (a\) ">\(\log _{a} 1=0\) \ (e\) ">\(\ln 1=0\)
    \ (a\) ">\(\log _{a} a=1\) \ (e\) ">\(\ln e=1\)
    Mali Inverse \ (a\) ">\(a^{\log _{a} x}=x\)
    \(\log _{a} a^{x}=x\)    		 \ (e\) ">\(e^{\ln x}=x\)
    \(\ln e^{x}=x\)
    Bidhaa Mali ya Logarithms \ (a\) ">\(\log _{a}(M \cdot N)=\log _{a} M+\log _{a} N\) \ (e\) ">\(\ln (M \cdot N)=\ln M+\ln N\)
    Mali ya Quotient ya Logarithms \ (a\) ">\(\log _{a} \frac{M}{N}=\log _{a} M-\log _{a} N\) \ (e\) ">\(\ln \frac{M}{N}=\ln M-\ln N\)
    Nguvu Mali ya Logarithms \ (a\) ">\(\log _{a} M^{p}=p \log _{a} M\) \ (e\) ">\(\ln M^{p}=p \ln M\)
    Jedwali 10.4.1
    • Mabadiliko-ya-msingi Mfumo
      Kwa misingi yoyote logarithmic\(a\) na\(b\), na\(M>0\),

      \(\begin{array}{ll}{\log _{a} M=\frac{\log _{b} M}{\log _{b} a}} & {\log _{a} M=\frac{\log M}{\log a}} & {\log _{a} M=\frac{\ln M}{\ln a}} \\ {\text { new base } b} & {\text { new base } 10} & {\text { new base } e}\end{array}\)