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Sinau Matematika “Pahami dengan mudah Logaritma | Matematika 10”

Published by Mustikawati on 22 Mei 2021

salam sehat dan bahagia..

yuuk sinau matematika…

semoga kita senantiasa diberi kesehatan, amin. Kali ini yang akan kita bahas adalah logaritma.

1. Pengertian Logaritma

Logartima merupakan invers dari bilangan berpangkat (eksponen). Secara umum ditulis:

$a^{c}=b\Leftrightarrow \textup{ }_{}^{a}\textrm{log}\textup{ }c$

dengan $a>0,a\neq 1,b>0″>, a disebut bilangan pokok logaritma atau basis, b disebut numerus, yaitu bilangan yang dilogaritmakan.</p> <h3 class=$ Contoh

Nyatakan bentuk berikut dalam bentuk logaritma.

a. 2³ = 8

b. $3^{-4}=\frac{1}{81}$

c. $10^{4}=10.000$

Penyelesaian:

a. 2³ = 8 $\Leftrightarrow _{}^{2}\textrm{log}\textup{ }8=3$

b. $3^{-4}=\frac{1}{81}\Leftrightarrow _{}^{3}\textrm{log}\textup{ }\frac{1}{81}=-4$

c. $10^{4}=10.000\Leftrightarrow _{}^{10}\textrm{log}\textup{ }10.000=4\textup{ atau }\textup{log}10.000=4$

2. Sifat-sifat Logaritma

Sifat-sifat logaritma berikut berlaku denga syarat

$p>0\textup{ dan }p\neq 1,a>0,b>0,\textup{ dan }n\epsilon R”>.</p> <figure class=$ Sifat-sifat logaritma

Contoh

1.Tentukan nilai berikut.

a. $_{}^{2}\textrm{log}\textup{ }24+_{}^{2}\textrm{log}\textup{ }3-_{}^{2}\textrm{log}\textup{ }9$

b. $2\cdot _{}^{2}\textrm{log}\textup{ }8+_{}^{2}\textrm{log}\textup{ }3-3\cdot _{}^{2}\textrm{log}\textup{ }\frac{1}{4}$

Penyelesaian:

a. $_{}^{2}\textrm{log}\textup{ }24+_{}^{2}\textrm{log}\textup{ }3-_{}^{2}\textrm{log}\textup{ }9=_{}^{2}\textrm{log}\textup{ }\left ( \frac{24\cdot 3}{9} \right )=_{}^{2}\textrm{log}\textup{ }8=3$

b. $2\cdot _{}^{2}\textrm{log}\textup{ }8+_{}^{2}\textrm{log}\textup{ }3-3\cdot _{}^{2}\textrm{log}\textup{ }\frac{1}{4}$

$=_{}^{2}\textrm{log}\textup{ }8^{2}+_{}^{2}\textrm{log}\textup{ }\sqrt{2}-_{}^{2}\textrm{log}\textup{ }\left (\frac{1}{4} \right )^{3}$ (sifat 3)

$=_{}^{2}\textrm{log}\textup{ }\left ( 2^{3} \right )^{2}+_{}^{2}\textrm{log}\textup{ }2^{\frac{1}{2}}-_{}^{2}\textrm{log}\textup{ }\left ( 2^{-2} \right )^{3}$

$=_{}^{2}\textrm{log}\textup{ }\left ( \frac{2^{6}\cdot 2^{\frac{1}{2}}}{2^{-6}} \right )=_{}^{2}\textrm{log}\textup{ }2^{6+\frac{1}{2}-\left ( -6 \right )}=_{}^{2}\textrm{log}\textup{ }2^{12\frac{1}{2}}$ (sifat 1 dan sifat 2)

$=2^{12\frac{1}{2}}\cdot _{}^{2}\textrm{log}\textup{ }2=12\frac{1}{2}$ (sifat 3 dan sifat 9)

2. Dengan menggunakan sifat logaritma, tentukan nilai berikut..

a. $_{}^{3}\textrm{log 27}$

b. $_{}^{\frac{1}{25}}\textrm{log 125}$

Penyelesaian:

a. $_{}^{3}\textrm{log 27}=_{}^{3}\textrm{log }3^{3}=3\cdot _{}^{3}\textrm{log 3}=1$ (sifat 3 dan sifat 9)

b. $_{}^{\frac{1}{25}}\textrm{log 125}=_{}^{5^{-2}}\textrm{log }5^{3}=\frac{3}{-2}_{}^{5}\textrm{log 5}=-\frac{3}{2}$ (sifat 6 dan sifat 9)

3. Jika diketahui log 2=0,3010 dan log 3=0,4771; tentukan nilai berikut.

a. log 12

b. log 0,125

Penyelesaian:

a. log 12=log(2.2.3)=log2+log2+log3=0,3010+0,3010+0,4771=1,0791

b. $\textup{log 0,125}=\textup{log }\frac{1}{8}=\textup{log }2^{-3}=-3\cdot \textup{log }2=-3\cdot 0,3010=-0,9030$

4. Tentukan nilai dari $_{}^{5}\textrm{log }7\cdot _{}^{7}\textrm{log 625}$ .

Penyelesaian:

$_{}^{5}\textrm{log }7\cdot _{}^{7}\textrm{log 625}=_{}^{5}\textrm{log }625=_{}^{5}\textrm{log }5^{4}=4$

5. Jika $_{}^{5}\textrm{log 4}=a\textup{ dan }_{}^{4}\textrm{log 3}=b$ , tentukan nilai dari $_{}^{3}\textrm{log 20}$ .

Penyelesaian:

Misalkan:

$_{}^{5}\textrm{log 4}=\frac{\textup{log 4}}{\textup{log 5}}=a\Leftrightarrow \textup{log 5}=\frac{\textup{log 4}}{\textup{a}}$

$_{}^{4}\textrm{log 3}=\frac{\textup{log 3}}{\textup{log 4}}=b\Leftrightarrow \textup{log 3}=b\cdot \textup{log 4}$

maka

$_{}^{3}\textrm{log 20}=\frac{\textup{log 20}}{\textup{log 3}}$

$=\frac{\textup{log }\left ( 4\cdot 5 \right )}{\textup{log 3}}$

$=\frac{\textup{log }4+\textup{log }5}{\textup{log 3}}$

$=\frac{\textup{log 4}+\frac{\textup{log 4}}{a}}{b\textup{ log a}}$

$=\frac{\textup{1}+\frac{\textup{1}}{a}}{b}=\frac{a+1}{ab}$

Published inMatematika SMA/SMK

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