Sinau Matematika “Ringkasan Bilangan Berpangkat Bentuk Akar dan Logaritma”

salam sehat dan bahagia..

yuuk sinau matematika…

Alhamdulilah puji syukur kita haturkan kepada Allah SWT yang senantiasa memberikan kita kesehatan dalam masa pandemi ini.

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Bilangan Berpangkat

Sifat-sifat Bilangan Berpangkat:

$a^{m}\cdot a^{n}=a^{m+n}$
$\frac{a^{m}}{a^{n}}=a^{m-n},m> n”></li><li><img decoding=$
$\left (a\cdot b \right )^{m}=a^{m}\cdot b^{m}$
$\left (\frac{a}{b} \right )^{m}=\frac{a^{m}}{b^{m}}$
$a^{0}=1$
$a^{-m}=\frac{1}{a^{m}}$
$a^{\frac{m}{n}}=\sqrt[n]{a^{m}},n\geq 2$

Bentuk Akar

Operasi pada bentuk akar.

1. Penjumlahan dan pengurangan

$a\sqrt{c}+b\sqrt{c}=\left ( a+b \right )\sqrt{c}$
$a\sqrt{c}-b\sqrt{c}=\left ( a-b \right )\sqrt{c}$

2. Perkalian bilangan real dengan bentuk akar

$a\cdot b\sqrt{c}=ab\sqrt{c}$

3. Perkalian bnetuk akar dengan bentuk akar

$\sqrt{a}\cdot \sqrt{b}=\sqrt{a\cdot b}$
$c\sqrt{d}\cdot e\sqrt{f}=ce\sqrt{df}$

4. Pembagian bentuk akar

$\frac{a}{\sqrt{b}}=\frac{a}{\sqrt{b}}\cdot \frac{\sqrt{b}}{\sqrt{b}}=\frac{a}{b}\sqrt{b}$
$\frac{c}{a+\sqrt{b}}=\frac{c}{a+\sqrt{b}}\cdot \frac{a-\sqrt{b}}{a-\sqrt{b}}=\frac{c\left ( \sqrt{a}-\sqrt{b} \right )}{a^{2}-b}$
$\frac{c}{\sqrt{a}+ \sqrt{b}}=\frac{c}{\sqrt{a}+ \sqrt{b}}\cdot \frac{\sqrt{a}- \sqrt{b}}{\sqrt{a}- \sqrt{b}}=\frac{c(\sqrt{a}- \sqrt{b})}{a-b}$

Logaritma

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

Sifat-sifat logaritma

$_{}^{p}\textrm{log(a.b)}=_{}^{p}\textrm{log(a)}+ _{}^{p}\textrm{log(b)}$
$_{}^{p}\textrm{log}\left (\frac{a}{b} \right )=_{}^{p}\textrm{log}\textup{ }a-_{}^{p}\textrm{log}\textup{ }b$
$_{}^{p}\textrm{log}\textup{ }a^{n}=n\cdot _{}^{p}\textrm{log}\textup{ }a$
$_{}^{a}\textrm{log}\textup{ }b=\frac{_{}^{p}\textrm{log}\textup{ }b}{_{}^{p}\textrm{log}\textup{ }a}$
$\frac{1}{_{}^{a}\textrm{log}\textup{ }b}=_{}^{b}\textrm{log}\textup{ }a$
$_{}^{a^{n}}\textrm{log}\textup{ }b^{m}=\frac{m}{n}\cdot _{}^{a}\textrm{log}\textup{ }b$
$a^{_{}^{a}\textrm{log}\textup{ }b}=b$
$_{}^{p}\textrm{log}\textup{ }1=0$
$_{}^{a}\textrm{log}\textup{ }a=1$
$_{}^{p}\textrm{log}\textup{ }a\cdot _{}^{a}\textrm{log}\textup{ }b=_{}^{p}\textrm{log}\textup{ }b$

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