Originally Posted by

**cmf0106**
Hey Guys I need help turning this into LATEX, I have tried at it for a good portion of the night but cant seem to find the correct combination.

Here is what I have so far if anyone would like to correct it.

10.(\frac{(x+8)^2}{x^4})^{-3}=\frac{(x^4)^3}{[(x+8)^2]^3}=

\frac{x^{12}

}{(x+8)^6}

11.(\frac{(x+3)^{-2}}{y^{-4}})^{-3}=\frac{[(x+3)^{-2}]^{-3}}{(y^{-4}0^{-3}}=\frac{(x+3)}^{(-2)}{(-3)}{y^{(-4)}{(-3)}}=\frac{(x+3)^6}{y^{12}}

*with [tex] tags removed of course*

Many thanks

You left out the part in red. When in the math tags, you would have had an error. Also, you had an extra } in the code somewhere...

You can use \bigg to make the parenthesis\brackets larger

so you could have

Code:

[tex]\bigg(\frac{(x+8)^2}{x^4}\bigg)^{-3}=\frac{(x^4)^3}{[(x+8)^2]^3}=
\frac{x^{12}}{(x+8)^6}[/tex]

which gives

$\displaystyle \bigg(\frac{(x+8)^2}{x^4}\bigg)^{-3}=\frac{(x^4)^3}{[(x+8)^2]^3}=

\frac{x^{12}}{(x+8)^6}$

You can make some of the same adjustments for the other one.

However I see some faults in the code for the second one.

(\frac{(x+3)^{-2}}{y^{-4}})^{-3}=\frac{[(x+3)^{-2}]^{-3}}{(y^{-4}0^{-3}}=\frac{(x+3)}^{(-2)}{(-3)}{y^{(-4)}{(-3)}}=\frac{(x+3)^6}{y^{12}}

That should be a parenthesis: )

There are a couple things that would give you a syntax error:

(\frac{(x+3)^{-2}}{y^{-4}})^{-3}=\frac{[(x+3)^{-2}]^{-3}}{(y^{-4}0^{-3}}=\frac{(x+3)}^{(-2)}{(-3)}{y^{(-4)}{(-3)}}=\frac{(x+3)^6}{y^{12}}

To get them in one exponent, just leave it as {(-2)(-3)} and {(-4)(-3)}.

The code should be [after making the larger parenthesis adjustment]:

\bigg(\frac{(x+3)^{-2}}{y^{-4}}\bigg)^{-3}=\frac{[(x+3)^{-2}]^{-3}}{(y^{-4})^{-3}}=\frac{(x+3)}^{(-2)(-3)}{y^{(-4)(-3)}}=\frac{(x+3)^6}{y^{12}}

However, there is still an error:

\bigg(\frac{(x+3)^{-2}}{y^{-4}}\bigg)^{-3}=\frac{[(x+3)^{-2}]^{-3}}{(y^{-4})^{-3}}=\frac{(x+3)}^{(-2)(-3)}{y^{(-4)}{(-3)}}=\frac{(x+3)^6}{y^{12}}

That brace is in the wrong place:

\bigg(\frac{(x+3)^{-2}}{y^{-4}}\bigg)^{-3}=\frac{[(x+3)^{-2}]^{-3}}{(y^{-4})^{-3}}=\frac{(x+3)}^{(-2)(-3)}}{y^{(-4)}{(-3)}}=\frac{(x+3)^6}{y^{12}}

The brace should be in the place where the blue one is.

So your code should now be:

\bigg(\frac{(x+3)^{-2}}{y^{-4}}\bigg)^{-3}=\frac{[(x+3)^{-2}]^{-3}}{(y^{-4})^{-3}}=\frac{(x+3)^{(-2)(-3)}}{y^{(-4)(-3)}}=\frac{(x+3)^6}{y^{12}}

which yields:

$\displaystyle \bigg(\frac{(x+3)^{-2}}{y^{-4}}\bigg)^{-3}=\frac{[(x+3)^{-2}]^{-3}}{(y^{-4})^{-3}}=\frac{(x+3)^{(-2)(-3)}}{y^{(-4)(-3)}}=\frac{(x+3)^6}{y^{12}}$

also, you can use \left( ... \right) for the parenthesis! Keep that in mind.

I hope this helps!

--Chris

EDIT: I see that o_O beat me again...I must be a slow typer