Ada 7 sifat pada logaritma ini yang akan membantu kamu dalam memecahkan masalah yang berkaitan dengan logaritma
Ada 7 sifat pada logaritma ini yang akan membantu kamu dalam memecahkan masalah yang berkaitan dengan logaritma yaitu :
Penjumlahan dan Pengurangan Sifat 1
a log x + a log y = a log x y ^a\log x + ^a\log y = ^a\log xy a log x + a log y = a log x y Contoh : Sederhanakanlah !
2 log 4 + 2 log 8 ^2\log 4 + 2\log 8 2 log 4 + 2 log 8 3 log 1 9 ^3\log \frac{1}{9} 3 log 9 1 + 3 log 81 ^3\log 81 3 log 81 2 log 2 2 ^2\log 2\sqrt{2} 2 log 2 2 + 2 log 4 2 ^2\log 4\sqrt{2} 2 log 4 2 Jawab :
2 log 4 + 2 log 8 = 2 log 4.8 = 2 log 32 = 5 ^2\log 4 + ^2\log 8 = ^2\log 4 . 8 = ^2\log 32 = 5 2 log 4 + 2 log 8 = 2 log 4.8 = 2 log 32 = 5 3 log 1 9 + 3 log 81 = 3 log 1 9 . 81 = 3 log 9 = 2 ^3\log \frac{1}{9} + ^3\log 81= ^3\log \frac{1}{9} . 81 = ^3\log 9 = 2 3 log 9 1 + 3 log 81 = 3 log 9 1 .81 = 3 log 9 = 2 2 log 2 2 + 2 log 4 2 = 2 log 2 2 . 4 2 = 2 log 16 = 4 ^2\log 2\sqrt{2} + ^2\log 4\sqrt{2}= ^2\log 2\sqrt{2} . 4\sqrt{2}= ^2\log 16 = 4 2 log 2 2 + 2 log 4 2 = 2 log 2 2 .4 2 = 2 log 16 = 4 Sifat 2
a log x – a log y = a log x y ^a\log x – ^a\log y = ^a\log\frac{x}{y} a log x – a log y = a log y x Contoh: Sederhanakanlah!
2 log 16 – 2 log 8 ^2\log 16 – ^2 \log 8 2 log 16 – 2 log 8 log 1.000 – log 100 \log 1.000 – \log 100 log 1.000– log 100 3 log 18 – 3 log 6 ^3\log 18 – ^3\log 6 3 log 18 – 3 log 6 Jawab :
2 log 16 – 2 log 8 = 2 log 16 8 = 2 log 2 = 1 ^2\log 16 – ^2 \log 8 = ^2\log \frac{16}{8} = 2\log 2 = 1 2 log 16 – 2 log 8 = 2 log 8 16 = 2 log 2 = 1 log 1.000 – log 100 = log 1000 100 = log 10 = 1 \log 1.000 – \log 100 = \log \frac{1000}{100} = \log 10 = 1 log 1.000– log 100 = log 100 1000 = log 10 = 1 3 log 18 – 3 log 6 = 3 log 18 6 = 1 ^3\log 18 – ^3\log 6 = ^3\log \frac{18}{6} = 1 3 log 18 – 3 log 6 = 3 log 6 18 = 1 Sifat 3
a log x n = n . a log x ^a\log x^n = n . ^a\log x a log x n = n . a log x Contoh : Sederhanakan!
2 log 3 + 4 log 3 2 \log 3 + 4 \log 3 2 log 3 + 4 log 3 2 log a + 2 log b 2 \log a + 2 \log b 2 log a + 2 log b Jawab:
solusi
2 log 3 + 4 log 3 = log 3 2 + log 3 4 = log 9 + log 81 = log 9.81 = log 729
\begin{equation}
\begin{split}
2 \log 3 + 4 \log 3 &= \log 3^2 + \log 3^4\\&= \log 9 + \log 81\\&= \log 9 . 81\\&= \log 729
\end{split}
\end{equation}
2 log 3 + 4 log 3 = log 3 2 + log 3 4 = log 9 + log 81 = log 9.81 = log 729 solusi
2 log a + 2 log b = log a 2 + log b 2 = log a 2 . b 2 = log ( a b ) 2
\begin{equation}
\begin{split}
2 \log a + 2 \log b &= \log a^2 + \log b^2\\&= \log a^2 . b^2\\&= \log (ab)^2
\end{split}
\end{equation}
2 log a + 2 log b = log a 2 + log b 2 = log a 2 . b 2 = log ( ab ) 2 Ingat :
log 2 x = log x . log x = ( log x ) 2 \log^2x = \log x . \log x = (\log x)^2 log 2 x = log x . log x = ( log x ) 2 log x 2 = 2 log x \log x^2 = 2 \log x log x 2 = 2 log x Jadi, log 2 x ≠ log x 2 \log^2x ≠ \log x^2 log 2 x = log x 2
log − 1 x = 1 log x \log^{-1}x = \frac{1}{\log x} log − 1 x = l o g x 1 log x − 1 = log 1 x = − log x \log x^{-1} = \log \frac{1}{x} = -\log x log x − 1 = log x 1 = − log x Jadi, log − 1 x ≠ log x − 1 \log ^{-1}x ≠ \log x^{-1} log − 1 x = log x − 1
Perbandingan Sifat 4 a log x = c log x c log a ^a\log x = \frac{{}^{c}\log x}{{}^{c}\log a} a log x = c l o g a c l o g x g log a = 1 a log g ^g\log a = \frac{1}{{}^{a}\log g} g log a = a l o g g 1 Contoh : Diketahui 5 log 3 = a ^5\log 3 = a 5 log 3 = a dan 3 log 4 = b ^3\log 4=b 3 log 4 = b . Nilai dari 4 log 15 = … ^4\log 15=… 4 log 15 = …
Solusi: 4 log 15 = 3 log 15 3 log 4 = 3 log ( 3 × 5 ) 3 log 4 = 3 log 3 + 3 log 5 3 log 4 = 1 + 1 a b × a a = a + 1 a b
\begin{equation}
\begin{split}
^4 \log 15 &= \frac{^3\log 15}{^3\log4} \\&= \frac{^3\log (3\times 5)}{^3\log4}\\&= \frac{^3\log 3+^3\log 5}{^3\log4}\\&= \frac{1+\frac{1}{a}}{b}\times \frac{a}{a}\\&= \frac{a+1}{ab}
\end{split}
\end{equation}
4 log 15 = 3 log 4 3 log 15 = 3 log 4 3 log ( 3 × 5 ) = 3 log 4 3 log 3 + 3 log 5 = b 1 + a 1 × a a = ab a + 1
Sifat 5 a log c = a log b ⋅ b log c ^a\log c=^a\log b \cdot ^b\log c a log c = a log b ⋅ b log c Contoh : Tentukan nilai dari 3 log 7 ⋅ 7 log 81 ^3\log 7 \cdot ^7\log 81 3 log 7 ⋅ 7 log 81 !
Solusi: 3 log 7 ⋅ 7 log 81 = 3 log 81 = 4
\begin{equation}
\begin{split}
^3\log 7 \cdot ^7\log 81&=^3\log 81 \\&= 4
\end{split}
\end{equation}
3 log 7 ⋅ 7 log 81 = 3 log 81 = 4
Sifat 6 Perhatikan uraian berikut untuk menunjukkan sifat 6 logaritma ini :
p n log a m = log a m log p n = m . log a n . log p = m n p log a {}^{{{p}^{n}}}\log {{a}^{m}}=\frac{\log {{a}^{m}}}{\log {{p}^{n}}}=\frac{m.\log a}{n.\log p}=\frac{m}{n}\ {}^{p}\log a p n log a m = l o g p n l o g a m = n . l o g p m . l o g a = n m p log a
Jika m = n m = n m = n maka diperoleh :p n log a m = log a n log p n = n . log a n . log p = p log a {}^{{{p}^{n}}}\log {{a}^{m}}=\frac{\log {{a}^{n}}}{\log {{p}^{n}}}=\frac{n.\log a}{n.\log p}=\ {}^{p}\log a p n log a m = l o g p n l o g a n = n . l o g p n . l o g a = p log a
Sehingga dapat disimpulkan bahwa : Untuk p dan a bilangan real positif p ≠ 1 maka :
p n log a m = m n p log a {}^{{{p}^{n}}}\log {{a}^{m}}=\frac{m}{n}\ {}^{p}\log a p n log a m = n m p log a
p n log a n = p log a {}^{{{p}^{n}}}\log {{a}^{n}}={}^{p}\log a p n log a n = p log a
Jika numerus dan bilangan pokok dipangkatkan dengan bilangan yang sama maka hasilnya tetap.
Contoh : Hitunglah !
8 log 16 ^8\log 16 8 log 16 8 log 64 ^8\log 64 8 log 64 Jika 3 log 5 = a ^3\log 5 = a 3 log 5 = a hitunglah 2 5 log 27 ^25\log 27 2 5 log 27 Jawab :
8 log 16 ^8\log 16 8 log 16
8 log 16 = 2 3 log 2 4 = 4 3 2 log 2 = 4 3 . 1 = 4 3
\begin{equation}
\begin{split}
8\log 16 &= ^{{{2}^{3}}}\log {{2}^{4}} \\&= \frac{4}{3}{{\ }^{2}}\log 2\\&= \frac{4}{3}.1 \\&= \frac{4}{3}
\end{split}
\end{equation}
8 log 16 = 2 3 log 2 4 = 3 4 2 log 2 = 3 4 .1 = 3 4 8 log 64 ^8\log 64 8 log 64
8 log 64 = 2 3 log 2 6 = 6 3 . 2 log 2 = 6 3 . 1 = 2
\begin{equation}
\begin{split}
^8\log 64&= {}^{{{2}^{3}}}\log {{2}^{6}}\\&=\frac{6}{3}.{}^{2}\log 2\\&=\frac{6}{3}.1\\&=2
\end{split}
\end{equation}
8 log 64 = 2 3 log 2 6 = 3 6 . 2 log 2 = 3 6 .1 = 2 3 log 5 = a ^3\log 5 = a 3 log 5 = a , maka :
\begin{equation}
\begin{split}
^25\log 27 &= {}^{{{5}^{2}}}\log {{3}^{3}}\\&=\frac{3}{2}{{.}^{5}}\log 3\\&=\frac{3}{2}.\frac{1}{{}^{3}\log 5}\\&=\frac{3}{2}.\frac{1}{a}\\&=\frac{3}{2a}
\end{split}
Sifat 7 Perhatikan uraian dibawah ini! Misalkan n = p log a n = ^p\log a n = p log a , maka a = p n a = p^n a = p n , oleh karena n = p log a n = ^p\log a n = p log a , maka p n = p p log a = a p^n = {{p}^{{}^{p}\log a}} = a p n = p p l o g a = a (karena a = p n a = p^n a = p n ) sehingga disimpulkan :
Untuk p dan a bilangan real p ≠ 1 maka
p p l o g a = a {{p}^{{}^{p}\\log a}} = a p p l o g a = a Contoh :
4 2 log 5 = ( 2 2 ) 2 log 5 {{4}^{{}^{2}\log 5}}={{\left( {{2}^{2}} \right)}^{{}^{2}\log 5}} 4 2 l o g 5 = ( 2 2 ) 2 l o g 5 3 3 log 2 = ( 3 1 / 2 ) 3 log 2 {{\sqrt{3}}^{{}^{3}\log 2}}={{\left( {{3}^{{}^{1}/{}_{2}}} \right)}^{{}^{3}\log 2}} 3 3 l o g 2 = ( 3 1 / 2 ) 3 l o g 2 Jawab :
4 2 log 5 = ( 2 2 ) 2 log 5 = ( 2 2 log 5 ) 2 = 5 2 = 25 {{4}^{{}^{2}\log 5}}={{\left( {{2}^{2}} \right)}^{{}^{2}\log 5}} ={{\left( {{2}^{{}^{2}\log 5}} \right)}^{2}} = 5^2 = 25 4 2 l o g 5 = ( 2 2 ) 2 l o g 5 = ( 2 2 l o g 5 ) 2 = 5 2 = 25 3 3 log 2 = ( 3 1 / 2 ) 3 log 2 = ( 3 3 log 2 ) 1 2 = 3 1 2 = 3 {{\sqrt{3}}^{{}^{3}\log 2}}={{\left( {{3}^{{}^{1}/{}_{2}}} \right)}^{{}^{3}\log 2}} = {{\left( {{3}^{{}^{3}\log 2}} \right)}^{\frac{1}{2}}}={{3}^{\frac{1}{2}}} =\sqrt{3} 3 3 l o g 2 = ( 3 1 / 2 ) 3 l o g 2 = ( 3 3 l o g 2 ) 2 1 = 3 2 1 = 3