just testing - Page 2

Drexel28 · #16 $Old$ November 20th, 2009, 12:17 AM

Quote:

Originally Posted by Soroban $View Post$

Captions?

. . ]

Hahahahahahaha, I am sorry. That was very clever.

nikk · #17 $Old$ November 20th, 2009, 02:33 AM

$1+1 = 4/2$

--------------------------
$1+2 = 3$

Soroban · #18 $Old$ November 20th, 2009, 06:05 AM

Hello, nikk!

Quote:

$1+1 = 4/2$

You can get a "prettier" result with \frac{4}{2}: . $\frac{4}{2}$

With practice you can create: . $\frac{\dfrac{1}{4} - \dfrac{1}{x^2}}{\dfrac{1}{2} - \dfrac{1}{x}} \;=\;\frac{x+2}{2x}$

nikk · #19 $Old$ November 20th, 2009, 09:09 AM

Quote:

Originally Posted by Soroban $View Post$

Hello, nikk!

You can get a "prettier" result with \frac{4}{2}: . $\frac{4}{2}$

With practice you can create: . $\frac{\dfrac{1}{4} - \dfrac{1}{x^2}}{\dfrac{1}{2} - \dfrac{1}{x}} \;=\;\frac{x+2}{2x}$

it look more prety

Paul46 · #20 $Old$ December 2nd, 2009, 10:29 AM

$\frac{1}{2}mv^2$

$20\:\log.\left(\frac{2x10^-1}{2x10^-5}\right)\:=\:80$

10\:\log.\left[\left(\frac{2x10^-1}^2{2x10^-5}^2\right)\right]\:=\:80

where did i go wrong on equation 3, i have deleted the math tags to show work, i was trying to get the large outside brackets [ ]

10 log.[(2x10^-1)^2/(2x10^-5)^2]

also say if i want to show the multiply sign should i use x or *

Opalg · #21 $Old$ December 2nd, 2009, 12:17 PM

Quote:

Originally Posted by Paul46 $View Post$

$\frac{1}{2}mv^2$

$20\:\log.\left(\frac{2x10^-1}{2x10^-5}\right)\:=\:80$

10\:\log.\left[\left(\frac{2x10^-1}^2{2x10^-5}^2\right)\right]\:=\:80

where did i go wrong on equation 3, i have deleted the math tags to show work, i was trying to get the large outside brackets [ ]

10 log.[(2x10^-1)^2/(2x10^-5)^2]

also say if i want to show the multiply sign should i use x or *

Either 10\:\log.\left[\left(\frac{2\times 10^{-1}}{2\times 10^{-5}}\right)^2\right]\:=\:80, giving $10\:\log.\left[\left(\frac{2\times 10^{-1}}{2\times 10^{-5}}\right)^2\right]\:=\:80$ ,

or 10\:\log.\left[\left(\frac{(2\times 10^{-1})^2}{(2\times 10^{-5})^2}\right)\right]\:=\:80, giving $10\:\log.\left[\left(\frac{(2\times 10^{-1})^2}{(2\times 10^{-5})^2}\right)\right]\:=\:80$ ,

will work. But you can't use braces { } to do the same job as parentheses ( ). They have completely different functions in TeX.

For the multiplication sign, use \times.

Notice also that if an exponent consists of more than one symbol then you need to enclose the symbols in braces. Otherwise only the first symbol will be superscripted. For example, 10^-1 gives $10^-1$ , but 10^{-1} gives $10^{-1}$ .

Paul46 · #22 $Old$ December 2nd, 2009, 12:21 PM

Thanks very very much! that is a massive help.

Paul46 · #23 $Old$ December 4th, 2009, 10:21 AM

I'll try this again:

I hope i get it right!

$10\:log.\left[\left(\frac{2\times10^{-1}}{2\times10^{-5}}\right)^3\right]\:=\:30\:log.\left(\frac{2\times10^{-1}}{2\times10^{-5}}\right)\:=\:80$
\:=\:80

Paul46 · #24 $Old$ December 4th, 2009, 10:24 AM

Why did the last part mess up with the \: & 80??

Paul46 · #25 $Old$ December 4th, 2009, 11:08 AM

I'll try again:

I hope i get better this time?!

10\:log.\:\left[\left(\frac{(2\times 10^{-1})^3}{2\times10^{-5})^3}\right)\right]\:=\:30\:log.\left(\frac{2\times10^{-1}){2\times10^{-5})\right)\:=120

Opalg · #26 $Old$ December 4th, 2009, 11:19 AM

Quote:

Originally Posted by Paul46 $View Post$

Why did the last part mess up with the \: & 80??

Probably because what you wrote was "[math]10\:log.\left[\left(\frac{2\times10^{-1}}{2\times10^{-5}}\right)^3\right]\:=\:30\:log.\left(\frac{2\times10^{-1}}{2\times10^{-5}}\right)\:=\:80[/math]\:=\:80", with an additional "\:=\:80" at the end.

While I'm about it, here are a couple of extra comments. You don't normally put a full stop after "log", and in TeX you should write "\log". The backslash converts the letters from italic to roman, which looks better (the same applies to other common functions: \cos, \sin, \exp, ...).

Also, you shouldn't normally need to put extra space in math expressions by inserting spaces like "\:". TeX is designed to incorporate good spacing automatically.