Difference between revisions of "User:Temperal/The Problem Solver's Resource2"

Revision as of 16:28, 29 September 2007

The Problem Solver's Resource

Introduction \| Other Tips and Tricks \| Methods of Proof \| You are currently viewing page 2.

This is just a quick review of logarithms and exponents; it's elementary content.

Exponentials: Do you really need this one?
Logarithms: If $b^a=x$ , $\log_b{x}=a$ . Note that a logarithm in base e, i.e. $\log_e{x}=a$ is notated as $\ln{x}=a$ , or the natural logarithm of x. If no base is specified, then a logarithm is assumed to be in base 10.

$a^x \cdot a^y=a^{x+y}$

$(a^x)^y=a^{xy}$

$\frac{a^x}{a^y}=a^{x-y}$

$a^0=1$ , where $a\ne 0$ .

$\log_b xy=\log_b x +\log_b y$

$\log_b x^y=y\cdot \log_b x$

$\log_b \frac{x}{y} =\log_b x-\log_b y$

$\log_b a=\frac{1}{\log_a b}$

$\log_b b=1$

$\log_b a=\frac{\log_x a}{\log_x b}$ , where x is a constant.

$\log_1 a$ and $\log_0 a$ are undefined.

@@ Line 15: / Line 15: @@
 <math>(a^x)^y=a^{xy}</math>
-<math>\frac{a^x}{a^y}</math>=a^{x-y}<math>
+<math>\frac{a^x}{a^y}=a^{x-y}</math>
-</math>a^0=1<math>, where </math>a\ne 0<math>.
+<math>a^0=1</math>, where <math>a\ne 0</math>.
-</math>\log_b xy=\log_b x +\log_b y <math>
+<math>\log_b xy=\log_b x +\log_b y </math>
-</math>\log_b x^y=y\cdot \log_b x <math>
+<math>\log_b x^y=y\cdot \log_b x </math>
-</math>\log_b \frac{x}{y} =\log_b x-\log_b y<math>
+<math>\log_b \frac{x}{y} =\log_b x-\log_b y</math>
-</math>\log_b a=\frac{1}{\log_a b}<math>
+<math>\log_b a=\frac{1}{\log_a b}</math>
-</math>\log_b b=1<math>
+<math>\log_b b=1</math>
-</math>\log_b a=\frac{\log_x a}{\log_x b}<math>, where x is a constant.
+<math>\log_b a=\frac{\log_x a}{\log_x b}</math>, where x is a constant.
-</math>\log_1 a<math> and </math>\log_0 a$ are undefined.
+<math>\log_1 a</math> and <math>\log_0 a</math> are undefined.