Mehlala ea Lipotso tse Buisanang ka Mesebetsi ea Logarithmic
Li-logarithm ke mohopolo oa bohlokoa lipalo, haholo-holo algebra le tlhahlobo. Li amana haufi-ufi le li-exponents 'me li sebelisoa khafetsa ho rarolla li-equation tsa exponential le lits'ebetsong tse fapaneng tsa saense le boenjiniere. Sengoloa sena se tla tšohla mathata a 'maloa a logarithm a atisang ho hlaha, hammoho le tlhaloso e felletseng ea bothata ka bong.
Selelekela sa Li-Logarithm
Li-logarithm ke tse fapaneng tsa li-exponents. Haeba re na le equation ea exponential \(b^y = x\), joale sebopeho sa eona sa logarithmic ke \(y = \log_b{x}\), e bolelang "y ke logarithm ea x e nang le motheo oa b". Tse ling tsa li-logarithm tse sebelisoang haholo ke logarithm ea tlhaho (base \(e\)) le logarithm ea decimal (base 10).
Matlotlo a Li-Logarithm
Tse latelang ke tse ling tsa litšobotsi tsa motheo tsa li-logarithm tse atisang ho sebelisoa ho rarolla mathata:
1. Logarithm ea sehlahisoa:
\[
\log_b{(xy)} = \log_b{x} + \log_b{y}
\]
2. Logarithm ea quotient:
\[
\log_b{(\frac{x}{y})} = \log_b{x} – \log_b{y}
\]
3. Logarithm ea exponent:
\[
\log_b{(x^a)} = a \cdot \log_b{x}
\]
4. Phetoho ea motheo oa logarithmic:
\[
\log_b{x} = \frac{\log_k{x}}{\log_k{b}}
\]
Lipotso tsa Mehlala le Puisano
1. Potso ea 1:
Fumana boleng ba \( \log_2{32} \).
Puisano:
Rea tseba hore \(32\) e ka ngolwa e le \(2^5\). Ka hona:
\[
\log_2{32} = \log_2{(2^5)} = 5 \cdot \log_2{2}
\]
Ho tloha ka \(\log_2{2} = 1\):
\[
\log_2{32} = 5 \cdot 1 = 5
\]
Kahoo, boleng ba \( \log_2{32} \) ke 5.
2. Potso ea 2:
Haeba \( \log_3{x} = 4 \), fumana boleng ba \( x \).
Puisano:
Ho latela tlhaloso ea logarithm, \( \log_3{x} = 4 \) e ka ngoloa bocha ka mokhoa oa exponential:
\[
3^4 = x
\]
Ho bala \(3^4\):
\[
3 ^ 4 = 81
\]
Kahoo, boleng ba \( x \) ke 81.
3. Potso ea 3:
Ho fanoe ka equation \( \log_{10}{x} = -2 \). Fumana boleng ba \( x \).
Puisano:
Fetolela foromo ea logarithmic ho ea ho foromo ea exponential:
\[
10^{-2} = x
\]
Ho bala \(10^{-2}\):
\[
10^{-2} = \frac{1}{10^2} = \frac{1}{100} = 0.01
\]
Kahoo, boleng ba \( x \) ke 0.01.
4. Potso ea 4:
Fumana boleng ba \( \log_5{(125 \cdot 25)} \).
Puisano:
Re tseba hore \(125 = 5^3\) le \(25 = 5^2\). Ebe:
\[
\log_5{(125 \cdot 25)} = \log_5{(5^3 \cdot 5^2)}
\]
Ho ipapisitsoe le thepa ea sehlahisoa sa li-logarithm:
\[
\log_5{(5^3 \cdot 5^2)} = \log_5{5^5}
\]
Ho sebelisa thepa ea matla a logarithmic:
\[
\log_5{5^5} = 5 \cdot \log_5{5}
\]
Ho tloha ka \(\log_5{5} = 1\):
\[
5 \cdot 1 = 5
\]
Kahoo, boleng ba \( \log_5{(125 \cdot 25)} \) ke 5.
5. Potso ea 5:
Fumana boleng ba \( \log_{2}{(8 \cdot \sqrt{2})} \).
Puisano:
Re tseba hore \(8 = 2^3\) le \(\sqrt{2} = 2^{1/2}\). Ebe:
\[
\log_{2}{(8 \cdot \sqrt{2})} = \log_{2}{(2^3 \cdot 2^{1/2})}
\]
Ho ipapisitsoe le thepa ea sehlahisoa sa li-logarithm:
\[
\log_{2}{(2^3 \cdot 2^{1/2})} = \log_{2}{(2^{3 + 1/2})} = \log_{2}{(2^{3.5})}
\]
Ho sebelisa thepa ea matla a logarithmic:
\[
\log_{2}{(2^{3.5})} = 3.5 \cdot \log_{2}{2}
\]
Ho tloha ka \(\log_{2}{2} = 1\):
\[
3.5 \cdot 1 = 3.5
\]
Kahoo, boleng ba \( \log_{2}{(8 \cdot \sqrt{2})} \) ke 3.5.
6. Potso ea 6:
Haeba \( \log_4{y} – \log_4{2} = 3 \), fumana boleng ba \( y \).
Puisano:
Ho ipapisitsoe le thepa ea logarithmic quotient:
\[
\log_4{(\frac{y}{2})} = 3
\]
Fetolela sebopeho sa logarithmic ho ba exponential:
\[
4^3 = \frac{y}{2}
\]
Ho bala \(4^3\):
\[
4 ^ 3 = 64
\]
Kahoo:
\[
64 = \frac{y}{2}
\]
Kahoo:
\[
y = 64 \cdot 2 = 128
\]
Kahoo, boleng ba \( y \) ke 128.
7. Potso ea 7:
Fumana boleng ba \( \log_{6}{\frac{1}{36}} \).
Puisano:
Re tseba hore \(36 = 6^2\). Ebe:
\[
\log_{6}{\frac{1}{36}} = \log_{6}{(6^{-2})}
\]
Ho sebelisa thepa ea matla a logarithmic:
\[
\log_{6}{(6^{-2})} = -2 \cdot \log_{6}{6}
\]
Ho tloha ka \(\log_{6}{6} = 1\):
\[
-2 \cdot 1 = -2
\]
Kahoo, boleng ba \( \log_{6}{\frac{1}{36}} \) ke -2.
Qetello
Li-logarithm ke sesebelisoa se thusang haholo sa lipalo lits'ebetsong tse fapaneng tsa saense le boenjiniere. Ho utloisisa litšobotsi tsa motheo tsa li-logarithm ho ka nolofatsa ho rarolla mathata a mangata. Sengoloa sena se hlalositse mathata a 'maloa' me sa tšohla li-logarithm tse hlahang khafetsa maemong a fapaneng. Ho itloaetsa le ho utloisisa likhopolo tsena ho tla thusa haholo ho tseba sehlooho sa li-logarithm.
