Artikel ini mengulas mengenai persamaan logaritma dan juga sifat-sifat yang digunakan untuk menyelesaikan soal-soal logaritma
Logaritma diperkenalkan pertama kali oleh John Napier (matematikawan Skotlandia). Napier mendapatkan suatu tata cara yang dipahami “Napierian Logarithm”. Sistem ini digunakan untuk perkiraan yang kompleks, tidak hanya melibatkan penjumlahan, pengurangan, perkalian dan pembagian, tetapi juga perpangkatan dan fungsi trigonometri.
Banyak sekali perkara dalam ilmu pengetahuan, teknologi maupun dalam kehidupan sehari-hari yang melibatkan fungsi atau persamaan logaritma, utamanya peristiwa kemajuan dan peluruhan. Hal ini dikarenakan logaritma ialah invers (kebalikan) dari eksponen.
Logaritma matematika juga digunakan untuk memecahkan perkara eksponen yang menyibukkan dicari akar-akar atau penyelesaiannya.
Persamaan Logaritma
Logaritma ialah invers (kebalikan) dari eksponen.
Bisa ditulis:
$a^{c}=b\ \Leftrightarrow\ _{}^{a}\textrm{log}\ b=c$
$a^{c}=b\ \Leftrightarrow\ _{}^{a}\textrm{log}\ b=c$
Logaritma dari b dengan bilangan pokok a ditulis selaku $_{}^{a}\textrm{log}\ b$ dengan a > 0, a ≠ 1 dan b ≥ 0 (b disebut numerus).
Suatu persamaan dengan numerus atau bilangan logaritmanya menampung variabel x disebut persamaan logaritma. Contoh persamaan logaritma:
$_{}^{2}\textrm{log}\ x\left ( x+4 \right )=\ _{}^{2}\textrm{log}\ 12$
$_{}^{3}\textrm{log}\ \left ( x+1 \right )+\ _{}^{3}\textrm{log}\ \left ( x+2 \right )=\ _{}^{9}\textrm{log}\ \left ( 4x+4 \right )$
Sifat-Sifat Logaritma
Sifat-sifat logaritma yaitu selaku berikut:
Misalkan q, a, b, n, p > 0, q ≠ 1 dan a ≠ 1, p ≠ 1:
A. (i) $_{}^{q}\textrm{log}\ 1 = 0$ (ii) $_{}^{q}\textrm{log}\ q = 1$ (iii) $_{}^{q}\textrm{log}\ q^{n} = n$
B. $_{}^{q}\textrm{log}\ ab=\ _{}^{q}\textrm{log}\ a+\ _{}^{q}\textrm{log}\ b$
C. $_{}^{q}\textrm{log}\ \frac{a}{b}=\ _{}^{q}\textrm{log}\ a-\ _{}^{q}\textrm{log}\ b$
D. $_{}^{q}\textrm{log}\ a^{n} = n\ _{}^{q}\textrm{log}\ a$
E. $_{}^{a}\textrm{log}\ b=\frac{_{}^{p}\textrm{log}\ a}{_{}^{p}\textrm{log}\ b}$
F. $a\ _{}^{a}\textrm{log}\ b=b$
G. $_{}^{a}\textrm{log}\ b\ \times\ _{}^{b}\textrm{log}\ a=1$
Catatan: Logaritma dengan bilangan pokok 10, bilangan pokok lazimnya tidak ditulis. Jadi, $_{}^{10}\textrm{log}\ a$ ditulis $_{}^{}\textrm{log}\ a$.
Contoh 1:
Sederhanakanlah bentuk logaritma berikut ini!
a. $3\ _{}^{3}\textrm{log}\ 5+2\ _{}^{3}\textrm{log}\ 2$
b. $\frac{1}{3}\ _{}^{5}\textrm{log}\ 27+ _{}^{5}\textrm{log}\ 3-\frac{2}{3}\ _{}^{5}\textrm{log}\ 3$
a. $3\ _{}^{3}\textrm{log}\ 5+2\ _{}^{3}\textrm{log}\ 2$
b. $\frac{1}{3}\ _{}^{5}\textrm{log}\ 27+ _{}^{5}\textrm{log}\ 3-\frac{2}{3}\ _{}^{5}\textrm{log}\ 3$
Jawab:
a. $3\ _{}^{3}\textrm{log}\ 5+2\ _{}^{3}\textrm{log}\ 2$
= $_{}^{3}\textrm{log}\ 5^{3}+_{}^{3}\textrm{log}\ 2^{2}$
= $_{}^{3}\textrm{log}\ 125+_{}^{3}\textrm{log}\ 4$
= $_{}^{3}\textrm{log}\ 125\ \times \ 4$
= $_{}^{3}\textrm{log}\ 500$
= $_{}^{3}\textrm{log}\ 5^{3}+_{}^{3}\textrm{log}\ 2^{2}$
= $_{}^{3}\textrm{log}\ 125+_{}^{3}\textrm{log}\ 4$
= $_{}^{3}\textrm{log}\ 125\ \times \ 4$
= $_{}^{3}\textrm{log}\ 500$
b. $\frac{1}{3}\ _{}^{5}\textrm{log}\ 27+ _{}^{5}\textrm{log}\ 3-\frac{2}{3}\ _{}^{5}\textrm{log}\ 3$
= $_{}^{5}\textrm{log}\ 27^{\frac{1}{3}}+ _{}^{5}\textrm{log}\ 3-_{}^{5}\textrm{log}\ 3^{\frac{2}{3}}$
= $_{}^{5}\textrm{log}\ \left ( 3^{3} \right )^{\frac{1}{3}}+ _{}^{5}\textrm{log}\ 3-_{}^{5}\textrm{log}\ 3^{\frac{2}{3}}$
= $_{}^{5}\textrm{log}\ 3+ _{}^{5}\textrm{log}\ \frac{3}{3^{\frac{2}{3}}}$
= $_{}^{5}\textrm{log}\ 3+ _{}^{5}\textrm{log}\ 3^{\frac{1}{3}}$
= $_{}^{5}\textrm{log}\ 3\ \times \ 3^{\frac{1}{3}}$
= $_{}^{5}\textrm{log}\ 3^{\frac{4}{3}}$
= $_{}^{5}\textrm{log}\ \sqrt[3]{81}$
= $_{}^{5}\textrm{log}\ 3\ \sqrt[3]{3}$
= $_{}^{5}\textrm{log}\ 27^{\frac{1}{3}}+ _{}^{5}\textrm{log}\ 3-_{}^{5}\textrm{log}\ 3^{\frac{2}{3}}$
= $_{}^{5}\textrm{log}\ \left ( 3^{3} \right )^{\frac{1}{3}}+ _{}^{5}\textrm{log}\ 3-_{}^{5}\textrm{log}\ 3^{\frac{2}{3}}$
= $_{}^{5}\textrm{log}\ 3+ _{}^{5}\textrm{log}\ \frac{3}{3^{\frac{2}{3}}}$
= $_{}^{5}\textrm{log}\ 3+ _{}^{5}\textrm{log}\ 3^{\frac{1}{3}}$
= $_{}^{5}\textrm{log}\ 3\ \times \ 3^{\frac{1}{3}}$
= $_{}^{5}\textrm{log}\ 3^{\frac{4}{3}}$
= $_{}^{5}\textrm{log}\ \sqrt[3]{81}$
= $_{}^{5}\textrm{log}\ 3\ \sqrt[3]{3}$
Contoh 2:
Tentukan nilai dari logaritma berikut ini!
a. $8\ _{}^{2}\textrm{log}\ 3$
b. $_{}^{b}\textrm{log}\ \frac{1}{a^{2}}\ \times \ _{}^{a}\textrm{log}\ b$
c. $\frac{1}{2}^{_{}^{2}\textrm{log}\ 4}\ \times \ 3^{_{}^{9}\textrm{log}\ 16}\ \times \ 5^{_{}^{\frac{1}{5}}\textrm{log}\ 2}$
Tentukan nilai dari logaritma berikut ini!
a. $8\ _{}^{2}\textrm{log}\ 3$
b. $_{}^{b}\textrm{log}\ \frac{1}{a^{2}}\ \times \ _{}^{a}\textrm{log}\ b$
c. $\frac{1}{2}^{_{}^{2}\textrm{log}\ 4}\ \times \ 3^{_{}^{9}\textrm{log}\ 16}\ \times \ 5^{_{}^{\frac{1}{5}}\textrm{log}\ 2}$
Jawab:
a. $8\ _{}^{2}\textrm{log}\ 3$
= $\left ( 2 \right )^{3}\ _{}^{2}\textrm{log}\ 3$
= $\left ( 2 \right )\ _{}^{2}\textrm{log}\ 3^{3}$
= $27$
= $\left ( 2 \right )^{3}\ _{}^{2}\textrm{log}\ 3$
= $\left ( 2 \right )\ _{}^{2}\textrm{log}\ 3^{3}$
= $27$
b. $_{}^{b}\textrm{log}\ \frac{1}{a^{2}}\ \times \ _{}^{a}\textrm{log}\ b$
= $_{}^{b}\textrm{log}\ a^{-2}\ \times \ _{}^{a}\textrm{log}\ b$
= $-2\ _{}^{b}\textrm{log}\ a \times \ _{}^{a}\textrm{log}\ b$
= $-2\ \times\ 1$
= $-2$
= $_{}^{b}\textrm{log}\ a^{-2}\ \times \ _{}^{a}\textrm{log}\ b$
= $-2\ _{}^{b}\textrm{log}\ a \times \ _{}^{a}\textrm{log}\ b$
= $-2\ \times\ 1$
= $-2$
c. $\frac{1}{2}^{_{}^{2}\textrm{log}\ 4}\ \times \ 3^{_{}^{9}\textrm{log}\ 16}\ \times \ 5^{_{}^{\frac{1}{5}}\textrm{log}\ 2}$
= $2^{_{}^-1\ \times\ _{}^{2}\textrm{log}\ 4}\ \times \ 9^{\frac{1}{2}\ \times\ _{}^{9}\textrm{log}\ 16}\ \times \ \frac{1}{5}^{-1\ \times\ _{}^{\frac{1}{5}}\textrm{log}\ 2}$
= $2^{_{}^{2}\textrm{log}\ 4^{-1}}\ \times \ 9^{_{}^{9}\textrm{log}\ 16^{\frac{1}{2}}}\ \times \ \frac{1}{5}^{_{}^{\frac{1}{5}}\textrm{log}\ 2^{-1}}$
= $4^{-1}\ \times\ 16^{\frac{1}{2}}\ \times \ 2^{-1}$
= $\frac{1}{4}\ \times\ 4\ \times\ \frac{1}{2}=\frac{1}{2}$
= $2^{_{}^-1\ \times\ _{}^{2}\textrm{log}\ 4}\ \times \ 9^{\frac{1}{2}\ \times\ _{}^{9}\textrm{log}\ 16}\ \times \ \frac{1}{5}^{-1\ \times\ _{}^{\frac{1}{5}}\textrm{log}\ 2}$
= $2^{_{}^{2}\textrm{log}\ 4^{-1}}\ \times \ 9^{_{}^{9}\textrm{log}\ 16^{\frac{1}{2}}}\ \times \ \frac{1}{5}^{_{}^{\frac{1}{5}}\textrm{log}\ 2^{-1}}$
= $4^{-1}\ \times\ 16^{\frac{1}{2}}\ \times \ 2^{-1}$
= $\frac{1}{4}\ \times\ 4\ \times\ \frac{1}{2}=\frac{1}{2}$
Menentukan Penyelesaian Persamaan Logaritma
Persamaan logaritma berupa $_{}^{a}\textrm{log}\ f\left ( x \right )\ =\ _{}^{a}\textrm{log}\ p$
Himpunan solusi dari persamaan logaritma $_{}^{a}\textrm{log}\ f\left ( x \right )\ =\ _{}^{a}\textrm{log}\ p$ dengan a > 0, a ≠ 1, sanggup diputuskan dengan sifat berikut:
Jika p > 0 dan $_{}^{a}\textrm{log}\ f\left ( x \right )\ =\ _{}^{a}\textrm{log}\ p$, maka $f\left ( x \right )\ =\ p$ asalkan $f\left ( x \right )\ > \ 0$
Contoh:
Tentukan himpunan solusi dari persamaan logaritma berikut!
a. $_{}^{}\textrm{log}\left ( 2x-3 \right )-_{}^{}\textrm{log}\left ( x-3 \right )=_{}^{}\textrm{log}\ 5$
b. $_{}^{3}\textrm{log}\ \left ( x-3 \right )\ =\ _{}^{9}\textrm{log}\ 16$
a. $_{}^{}\textrm{log}\left ( 2x-3 \right )-_{}^{}\textrm{log}\left ( x-3 \right )=_{}^{}\textrm{log}\ 5$
b. $_{}^{3}\textrm{log}\ \left ( x-3 \right )\ =\ _{}^{9}\textrm{log}\ 16$
Jawab:
a. $_{}^{}\textrm{log}\left ( 2x-3 \right )-_{}^{}\textrm{log}\left ( x-3 \right )=_{}^{}\textrm{log}\ 5$
(i) Numerus mesti nyata $\left ( f\left ( x \right )\ > \ 0 \right )$
$2x-3> 0\ \Leftrightarrow\ x> \frac{3}{2}$
$x-3> 0\ \Leftrightarrow\ x> 3$
Syarat numerus yang mesti dipenuhi yaitu $x> 3$
(i) Numerus mesti nyata $\left ( f\left ( x \right )\ > \ 0 \right )$
$2x-3> 0\ \Leftrightarrow\ x> \frac{3}{2}$
$x-3> 0\ \Leftrightarrow\ x> 3$
Syarat numerus yang mesti dipenuhi yaitu $x> 3$
(ii) $_{}^{}\textrm{log}\left ( 2x-3 \right )-_{}^{}\textrm{log}\left ( x-3 \right )=_{}^{}\textrm{log}\ 5$
$_{}^{}\textrm{log}\frac{2x-3}{x-3}=_{}^{}\textrm{log}\ 5$
$\frac{2x-3}{x-3}=5$
$2x-3=5x-15$
$3x=12$
$x=4$
$_{}^{}\textrm{log}\frac{2x-3}{x-3}=_{}^{}\textrm{log}\ 5$
$\frac{2x-3}{x-3}=5$
$2x-3=5x-15$
$3x=12$
$x=4$
Karena $x> 3$, maka himpunan penyelesaiannya yaitu {4}.
b. $_{}^{3}\textrm{log}\ \left ( x-3 \right )\ =\ _{}^{9}\textrm{log}\ 16$
(i) Numerus mesti nyata $\left ( f\left ( x \right )\ > \ 0 \right )$
$x-3> 0\ \Leftrightarrow\ x> 3$
Syarat numerus yang mesti dipenuhi yaitu $x> 3$
$x-3> 0\ \Leftrightarrow\ x> 3$
Syarat numerus yang mesti dipenuhi yaitu $x> 3$
(ii) $_{}^{3}\textrm{log}\ \left ( x-3 \right )\ =\ _{}^{9}\textrm{log}\ 16$
$_{}^{3}\textrm{log}\ \left ( x-3 \right )\ =\ \frac{_{}^{3}\textrm{log}\ 16}{_{}^{3}\textrm{log}\ 9}$
$_{}^{3}\textrm{log}\ \left ( x-3 \right )\ =\ \frac{_{}^{3}\textrm{log}\ 16}{2}$
$2\ \times\ _{}^{3}\textrm{log}\ \left ( x-3 \right )\ =\ _{}^{3}\textrm{log}\ 16$
$_{}^{3}\textrm{log}\ \left ( x-3 \right )^{2}\ =\ _{}^{3}\textrm{log}\ 16$
$\left ( x-3 \right )^{2}\ =\ 16$
$x^{2}-6x+9=16$
$x^{2}-6x-7=0$
$\left ( x-7 \right )\left ( x+1 \right )=0$
$x=7$ atau $x=-1$
$_{}^{3}\textrm{log}\ \left ( x-3 \right )\ =\ \frac{_{}^{3}\textrm{log}\ 16}{_{}^{3}\textrm{log}\ 9}$
$_{}^{3}\textrm{log}\ \left ( x-3 \right )\ =\ \frac{_{}^{3}\textrm{log}\ 16}{2}$
$2\ \times\ _{}^{3}\textrm{log}\ \left ( x-3 \right )\ =\ _{}^{3}\textrm{log}\ 16$
$_{}^{3}\textrm{log}\ \left ( x-3 \right )^{2}\ =\ _{}^{3}\textrm{log}\ 16$
$\left ( x-3 \right )^{2}\ =\ 16$
$x^{2}-6x+9=16$
$x^{2}-6x-7=0$
$\left ( x-7 \right )\left ( x+1 \right )=0$
$x=7$ atau $x=-1$
Karena $x> 3$, maka himpunan penyelesaiannya yaitu {7}.
Persamaan logaritma berupa $_{}^{a}\textrm{log}\ f\left ( x \right )\ =\ _{}^{b}\textrm{log}\ f\left ( x \right )$
Himpunan solusi dari persamaan logaritma $_{}^{a}\textrm{log}\ f\left ( x \right )\ =\ _{}^{b}\textrm{log}\ f\left ( x \right )$ dengan a > 0, a ≠ b serta $f\left ( x \right )$ dan $g\left ( x \right )$ fungsi aljabar, sanggup diputuskan dengan sifat berikut:
Jika $_{}^{a}\textrm{log}\ f\left ( x \right )\ =\ _{}^{b}\textrm{log}\ f\left ( x \right )$, maka $f\left ( x \right )\ =\ 1$ asalkan $f\left ( x \right )\ > \ 0$, a ≠ b
Persamaan logaritma berupa $_{}^{a}\textrm{log}\ f\left ( x \right )\ =\ _{}^{a}\textrm{log}\ g\left ( x \right )$
Himpunan solusi dari persamaan logaritma $_{}^{a}\textrm{log}\ f\left ( x \right )\ =\ _{}^{a}\textrm{log}\ g\left ( x \right )$ dengan a > 0, a ≠ 1 serta $f\left ( x \right )$ dan $g\left ( x \right )$ fungsi aljabar, sanggup diputuskan dengan sifat berikut:
Jika $_{}^{a}\textrm{log}\ f\left ( x \right )\ =\ _{}^{a}\textrm{log}\ g\left ( x \right )$, maka $f\left ( x \right )=g\left ( x \right )$ asalkan $f\left ( x \right )> 0,\ g\left ( x \right )> 0$
Contoh:
Tentukan himpunan solusi dari persamaan logaritma berikut!
a. $_{}^{3}\textrm{log}\left ( x^{2}+3x+2 \right )=\ _{}^{3}\textrm{log}\left ( 5x+5 \right )$
b. $_{}^{2}\textrm{log}\left ( _{}^{2}\textrm{log}\left ( x+7 \right )+1 \right )=\ _{}^{2}\textrm{log}\left ( _{}^{2}\textrm{log}\ x\ +\ _{}^{2}\textrm{log}\left ( x-3 \right ) \right )$
a. $_{}^{3}\textrm{log}\left ( x^{2}+3x+2 \right )=\ _{}^{3}\textrm{log}\left ( 5x+5 \right )$
b. $_{}^{2}\textrm{log}\left ( _{}^{2}\textrm{log}\left ( x+7 \right )+1 \right )=\ _{}^{2}\textrm{log}\left ( _{}^{2}\textrm{log}\ x\ +\ _{}^{2}\textrm{log}\left ( x-3 \right ) \right )$
Jawab:
a. $_{}^{3}\textrm{log}\left ( x^{2}+3x+2 \right )=\ _{}^{3}\textrm{log}\left ( 5x+5 \right )$
(i) Numerus mesti nyata $\left ( f\left ( x \right )\ > \ 0 \right )$
$\Leftrightarrow\ x^{2}+3x+2> 0\ \Leftrightarrow \ \left ( x+2 \right )\left ( x+1 \right )> 0$
Jadi, $x< -2$ atau $x>-1$
$\Leftrightarrow\ 5x+5> 0\ \Leftrightarrow \ x> -1$
Syarat numerus yang mesti dipenuhi yaitu $x> -1$
(i) Numerus mesti nyata $\left ( f\left ( x \right )\ > \ 0 \right )$
$\Leftrightarrow\ x^{2}+3x+2> 0\ \Leftrightarrow \ \left ( x+2 \right )\left ( x+1 \right )> 0$
Jadi, $x< -2$ atau $x>-1$
$\Leftrightarrow\ 5x+5> 0\ \Leftrightarrow \ x> -1$
Syarat numerus yang mesti dipenuhi yaitu $x> -1$
(ii) $_{}^{3}\textrm{log}\left ( x^{2}+3x+2 \right )=\ _{}^{3}\textrm{log}\left ( 5x+5 \right )$
$\Leftrightarrow\ x^{2}+3x+2=5x+5$
$\Leftrightarrow\ x^{2}-2x-3=0$
$\Leftrightarrow\ \left ( x-3 \right )\left ( x+1 \right )=0$
$\Leftrightarrow\ x=3$ atau $x=-1$
Karena $x> -1$, maka himpunan penyelesaiannya yaitu {3}.
$\Leftrightarrow\ x^{2}+3x+2=5x+5$
$\Leftrightarrow\ x^{2}-2x-3=0$
$\Leftrightarrow\ \left ( x-3 \right )\left ( x+1 \right )=0$
$\Leftrightarrow\ x=3$ atau $x=-1$
Karena $x> -1$, maka himpunan penyelesaiannya yaitu {3}.
b. Persamaan ini berisikan dua logaritma, jadi kita tuntaskan satu persatu.
$_{}^{2}\textrm{log}\left ( _{}^{2}\textrm{log}\left ( x+7 \right )+1 \right )=\ _{}^{2}\textrm{log}\left ( _{}^{2}\textrm{log}\ x\ +\ _{}^{2}\textrm{log}\left ( x-3 \right ) \right )$
$\Leftrightarrow\ _{}^{2}\textrm{log}\left ( _{}^{2}\textrm{log}\left ( x+7 \right )+\ _{}^{2}\textrm{log}\ 2 \right )=\ _{}^{2}\textrm{log}\left ( _{}^{2}\textrm{log}\ x\left ( x-3 \right ) \right )$
$\Leftrightarrow\ _{}^{2}\textrm{log}\left ( _{}^{2}\textrm{log}\ 2\left ( x+7 \right ) \right )=\ _{}^{2}\textrm{log}\left ( _{}^{2}\textrm{log}\ x\left ( x-3 \right ) \right )$
$_{}^{2}\textrm{log}\left ( _{}^{2}\textrm{log}\left ( x+7 \right )+1 \right )=\ _{}^{2}\textrm{log}\left ( _{}^{2}\textrm{log}\ x\ +\ _{}^{2}\textrm{log}\left ( x-3 \right ) \right )$
$\Leftrightarrow\ _{}^{2}\textrm{log}\left ( _{}^{2}\textrm{log}\left ( x+7 \right )+\ _{}^{2}\textrm{log}\ 2 \right )=\ _{}^{2}\textrm{log}\left ( _{}^{2}\textrm{log}\ x\left ( x-3 \right ) \right )$
$\Leftrightarrow\ _{}^{2}\textrm{log}\left ( _{}^{2}\textrm{log}\ 2\left ( x+7 \right ) \right )=\ _{}^{2}\textrm{log}\left ( _{}^{2}\textrm{log}\ x\left ( x-3 \right ) \right )$
Dengan meniadakan logaritma yang pertama, maka menjadi:
$_{}^{2}\textrm{log}\ \left ( 2x+14 \right )=\ _{}^{2}\textrm{log}\ x\left ( x^{2}-3x \right )$
$_{}^{2}\textrm{log}\ \left ( 2x+14 \right )=\ _{}^{2}\textrm{log}\ x\left ( x^{2}-3x \right )$
(i) Numerus mesti nyata $\left ( f\left ( x \right )\ > \ 0 \right )$
$2x+14> 0\ \Leftrightarrow\ x> -7\\ x^{2}-3x> 0\ \Leftrightarrow\ x\left ( x-3 \right )> 0$
$\Leftrightarrow\ x< 0$ atau $x>3$
Syarat numerus yang mesti dipenuhi yaitu $x>3$
$2x+14> 0\ \Leftrightarrow\ x> -7\\ x^{2}-3x> 0\ \Leftrightarrow\ x\left ( x-3 \right )> 0$
$\Leftrightarrow\ x< 0$ atau $x>3$
Syarat numerus yang mesti dipenuhi yaitu $x>3$
(ii) $_{}^{2}\textrm{log}\ \left ( 2x+14 \right )=\ _{}^{2}\textrm{log}\ x\left ( x^{2}-3x \right )$
$\Leftrightarrow\ 2x+14=x^{2}-3x\\ \Leftrightarrow\ x^{2}-5x-14=0\\ \Leftrightarrow\ \left ( x-7 \right )\left ( x+2 \right )=0$
$\Leftrightarrow\ x=7$ atau $x=-2$
Karena $x>3$, maka himpunan penyelesaiannya yaitu {7}.
$\Leftrightarrow\ 2x+14=x^{2}-3x\\ \Leftrightarrow\ x^{2}-5x-14=0\\ \Leftrightarrow\ \left ( x-7 \right )\left ( x+2 \right )=0$
$\Leftrightarrow\ x=7$ atau $x=-2$
Karena $x>3$, maka himpunan penyelesaiannya yaitu {7}.
Persamaan logaritma berupa $_{}^{h\left ( x \right )}\textrm{log}\ f\left ( x \right )\ =\ _{}^{h\left ( x \right )}\textrm{log}\ g\left ( x \right )$
Himpunan solusi dari persamaan logaritma $_{}^{h\left ( x \right )}\textrm{log}\ f\left ( x \right )\ =\ _{}^{h\left ( x \right )}\textrm{log}\ g\left ( x \right )$ dengan $f\left ( x \right )$, $g\left ( x \right )$ dan $h\left ( x \right )$ fungsi aljabar, sanggup diputuskan dengan sifat berikut:
Jika $_{}^{h\left ( x \right )}\textrm{log}\ f\left ( x \right )\ =\ _{}^{h\left ( x \right )}\textrm{log}\ g\left ( x \right )$, maka $f\left ( x \right )=g\left ( x \right )$ asalkan $f\left ( x \right )> 0,\ g\left ( x \right )> 0$ serta $h\left ( x \right )> 0$ dan $h\left ( x \right )\neq 1$
Contoh:
Tentukan himpunan solusi dari persamaan logaritma berikut!
$_{}^{x}\textrm{log}\left ( x+15 \right )\ -\ 2\ _{}^{x}\textrm{log}\ 10\ +\ 1=0$
$_{}^{x}\textrm{log}\left ( x+15 \right )\ -\ 2\ _{}^{x}\textrm{log}\ 10\ +\ 1=0$
Jawab:
$_{}^{x}\textrm{log}\left ( x+15 \right )\ -\ 2\ _{}^{x}\textrm{log}\ 10\ +\ _{}^{x}\textrm{log}\ x=0\\ \Leftrightarrow\ _{}^{x}\textrm{log}\left ( x+15 \right )\ +\ _{}^{x}\textrm{log}\ x=\ 2\ _{}^{x}\textrm{log}\ 10\\ \Leftrightarrow\ _{}^{x}\textrm{log}\ x\left ( x+15 \right )=\ _{}^{x}\textrm{log}\ 10^{2}\\ \Leftrightarrow\ _{}^{x}\textrm{log}\ \left ( x^{2}+15x \right )=\ _{}^{x}\textrm{log}\ 100$
(i) Numerus dan bilangan pokok
$h\left ( x \right )> 0$ dan $h\left ( x \right )\neq 1\ \Leftrightarrow\ x> 0$ dan $x\neq 1$
$f\left ( x \right )> 0\ \Leftrightarrow\ x^{2}\ +\ 15x> 0\\ \Leftrightarrow\ x\left ( x+15 \right )> 0$
$\Leftrightarrow\ x< -15$ atau $x>0$
Syarat bilangan pokok dan numerus yang mesti dipenuhi yaitu $x>0$ dan $x\neq 1$.
$h\left ( x \right )> 0$ dan $h\left ( x \right )\neq 1\ \Leftrightarrow\ x> 0$ dan $x\neq 1$
$f\left ( x \right )> 0\ \Leftrightarrow\ x^{2}\ +\ 15x> 0\\ \Leftrightarrow\ x\left ( x+15 \right )> 0$
$\Leftrightarrow\ x< -15$ atau $x>0$
Syarat bilangan pokok dan numerus yang mesti dipenuhi yaitu $x>0$ dan $x\neq 1$.
(ii) $_{}^{x}\textrm{log}\ \left ( x^{2}+15x \right )=\ _{}^{x}\textrm{log}\ 100$
$\Leftrightarrow\ x^{2}+15x=100\\ \Leftrightarrow\ x^{2}+15x-100=0\\ \Leftrightarrow\ \left ( x+20 \right )\left ( x-5 \right )=0$
$\Leftrightarrow\ x=-20$ atau $x=5$
Karena $x>0$ dan $x\neq 1$, maka himpunan penyelesaiannya yaitu {5}.
$\Leftrightarrow\ x^{2}+15x=100\\ \Leftrightarrow\ x^{2}+15x-100=0\\ \Leftrightarrow\ \left ( x+20 \right )\left ( x-5 \right )=0$
$\Leftrightarrow\ x=-20$ atau $x=5$
Karena $x>0$ dan $x\neq 1$, maka himpunan penyelesaiannya yaitu {5}.
Persamaan logaritma berupa $A\left \{ _{}^{a}\textrm{log}\ f\left ( x \right ) \right \}^{2}\ +\ B\left \{ _{}^{a}\textrm{log}\ f\left ( x \right ) \right \}\ +\ C=0$
Himpunan solusi dari persamaan logaritma $A\left \{ _{}^{a}\textrm{log}\ f\left ( x \right ) \right \}^{2}\ +\ B\left \{ _{}^{a}\textrm{log}\ f\left ( x \right ) \right \}\ +\ C=0$ dengan $a> 0$, $a\neq 1$ serta $f\left ( x \right )> 0$, sanggup diputuskan dengan merubah persamaan tersebut ke dalam bentuk persamaan kuadrat.
Contoh:
Tentukan himpunan solusi dari persamaan logaritma berikut!
$_{}^{2}\textrm{log}^{2}\ x\ -\ 2\ _{}^{2}\textrm{log}\ x^{2}=5$
$_{}^{2}\textrm{log}^{2}\ x\ -\ 2\ _{}^{2}\textrm{log}\ x^{2}=5$
Jawab:
$_{}^{2}\textrm{log}^{2}\ x\ -\ 2\ _{}^{2}\textrm{log}\ x^{2}=5\\ \Leftrightarrow\ \left ( _{}^{2}\textrm{log}\ x \right )^{2}-\ 4\ _{}^{2}\textrm{log}\ x\ -\ 5=0\\ \Leftrightarrow\ p^{2}-4p-5=0\\ \Leftrightarrow\ \left ( p-5 \right )\left ( p+1 \right )=0$
$\Leftrightarrow\ p=5$ atau $p=-1$
$\Leftrightarrow\ p=5$ atau $p=-1$
Untuk $p=-1$
$\Leftrightarrow\ _{}^{2}\textrm{log}\ x=5\\ \Leftrightarrow\ x=2^{5}\\ \Leftrightarrow\ x=32$
$\Leftrightarrow\ _{}^{2}\textrm{log}\ x=5\\ \Leftrightarrow\ x=2^{5}\\ \Leftrightarrow\ x=32$
Untuk $p=5$
$\Leftrightarrow\ _{}^{2}\textrm{log}\ x=-1\\ \Leftrightarrow\ x=2^{-1}\\ \Leftrightarrow\ x=\frac{1}{2}$
$\Leftrightarrow\ _{}^{2}\textrm{log}\ x=-1\\ \Leftrightarrow\ x=2^{-1}\\ \Leftrightarrow\ x=\frac{1}{2}$
Jadi, himpunan solusi persamaan tersebut yaitu $\left \{ \frac{1}{2},32 \right \}$.
Demikian ulasan bahan persamaan dan sifat logaritma, biar bermanfaat. Bila teman-teman memerlukan file pdf postingan ini sanggup sobat unduh lewat tombol berikut: