# Thread: Use $$and$$ tags NOT $$and$$ tags for latex.

1. ## Re: Use $$and$$ tags NOT $$and$$ tags for latex.

$$\sqrt {{b^2} - 4ac}$$

2. ## Re: Use $$and$$ tags NOT $$and$$ tags for latex.

am quite new to this forum and i have little idea on how to use the editor but please this is my question
Please am stuck with the integration I kindly need assistance in solving this equation, I will really appreciate if anyone could help with the solution or a guide on what i should do

$$C(N)=\int_{0}^{\infty}log(1+\frac{\rho}{a}z)\frac{ N\left(e^{-z}z^{y-1}\right)}{\left(y-1\right)!}\left(1-e^{-z}\sum_{i=0}^{y-1}\frac{z^{i}}{i!}\right)^{N-1}dz[\tex] 3. The closing tag should be [/TEX], not [\TEX]. $C(N)=\int_{0}^{\infty}log(1+\frac{\rho}{a}z)\frac{ N\left(e^{-z}z^{y-1}\right)}{\left(y-1\right)!}\left(1-e^{-z}\sum_{i=0}^{y-1}\frac{z^{i}}{i!}\right)^{N-1}dz$ Also it is better to write \log rather than log to indicate that it is an operator name rather than the product of three variables l,o,g. $C(N)=\int_{0}^{\infty}\log(1+\frac{\rho}{a}z)\frac { N\left(e^{-z}z^{y-1}\right)}{\left(y-1\right)!}\left(1-e^{-z}\sum_{i=0}^{y-1}\frac{z^{i}}{i!}\right)^{N-1}dz$ 4. ## Re: Use [tex] and$$ tags NOT $$and$$ tags for latex.

[/Tex]
\int_1^e \dfrac{1}{x}dx=lne-ln1=1
[/Tex]

5. ## Re: Use $$and$$ tags NOT $$and$$ tags for latex.

Originally Posted by Loser66
[/Tex]
\int_1^e \dfrac{1}{x}dx=lne-ln1=1
[/Tex]
Don't use the / in the opening delimiter. Should be this to begin: [TEX]. So your equation is this:

$\int_1^e \dfrac{1}{x}dx=lne-ln1=1$

7. ## Re: Use $$and$$ tags NOT $$and$$ tags for latex.

Originally Posted by Mr.MathType
Don't use the / in the opening delimiter. Should be this to begin: [TEX]. So your equation is this:

$\int_1^e \dfrac{1}{x}dx=lne-ln1=1$
\$\int_1^e \dfrac{1}{x}dx=lne-ln1=1\$ using the dollar sign gives a better rendering.

$\int_1^e \dfrac{1}{x}dx=lne-ln1=1$

8. ## Re: Use $$and$$ tags NOT $$and$$ tags for latex.

Actually I just noticed something else that would make it look better: Code the log as \ln, not simply ln. So now we have either $\int_1^e {\frac{1}{x}dx} = \ln e - \ln 1 = 1$ or

$\int_1^e {\frac{1}{x}dx} = \ln e - \ln 1 = 1$,

depending on whether it's inline or not.

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