khanpatel321
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How do I go deal with the x

4/x ( 4lnx + 3) - 4/x (4lnx - 3) all over (4lnx + 3 )^2

Thank you
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Smaug123
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(Original post by khanpatel321)
How do I go deal with the x

4/x ( 4lnx + 3) - 4/x (4lnx - 3) all over (4lnx + 3 )^2

Thank you
\dfrac{\frac{4}{x} (4 \log(x)+3) - \frac{4}{x} (4 \log(x) - 3)}{(4 \log(x)+3)^2}?
If that's the case, the numerator looks like ab-ac, and you can factorise it out.
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TenOfThem
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(Original post by khanpatel321)
How do I go deal with the x

4/x ( 4lnx + 3) - 4/x (4lnx - 3) all over (4lnx + 3 )^2

Thank you
You have \dfrac{4(4\ln (x)+3-4\ln (x)-3)}{x(4\ln (x)+3)^2}


and the numerator will simplify quite a lot
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khanpatel321
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(Original post by Smaug123)
\dfrac{\frac{4}{x} (4 \log(x)+3) - \frac{4}{x} (4 \log(x) - 3)}{(4 \log(x)+3)^2}?
If that's the case, the numerator looks like ab-ac, and you can factorise it out.
(Original post by TenOfThem)
You have \dfrac{4(4\ln (x)+3-4\ln (x)-3)}{x(4\ln (x)+3)^2}


and the numerator will simplify quite a lot

If instead of the x there was a number could you do the same thing?

So it would be
\dfrac{4(4\ln (x)+3-4\ln (x)-3)}{NUMBER(4\ln (x)+3)^2}
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TenOfThem
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(Original post by khanpatel321)
If instead of the x there was a number could you do the same thing?

So it would be
\dfrac{4(4\ln (x)+3-4\ln (x)-3)}{NUMBER(4\ln (x)+3)^2}
Sure, it is a common factor
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Davelittle
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(Original post by TenOfThem)
You have \dfrac{4(4\ln (x)+3-4\ln (x)-3)}{x(4\ln (x)+3)^2}


and the numerator will simplify quite a lot
Is the numerator not just 0?
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user2020user
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(Original post by TenOfThem)
You have \dfrac{4(4\ln (x)+3-4\ln (x)-3)}{x(4\ln (x)+3)^2}

and the numerator will simplify quite a lot
I got your back Tenten!

(Original post by Davelittle)
Is the numerator not just 0?
Nope. TenOfThem made a mistake with the simplification by overlooking the parentheses:

\begin{aligned} \sigma & = \dfrac{4}{x} \left( 4\log(x) + 3 \right) - \dfrac{4}{x} \left( 4\log(x) - 3 \right) \\ & = \dfrac{4}{x} \Big[ \left( 4\log(x) + 3 \right) - \left( 4\log(x) - 3 \right) \Big] \\ & = \ ... \end{aligned}
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TenOfThem
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(Original post by Khallil)
I got your back Tenten!



Nope. TenOfThem made a mistake with the simplification by overlooking the parentheses:

\begin{aligned} \sigma & = \dfrac{4}{x} \left( 4\log(x) + 3 \right) - \dfrac{4}{x} \left( 4\log(x) - 3 \right) \\ & = \dfrac{4}{x} \Big[ \left( 4\log(x) + 3 \right) - \left( 4\log(x) - 3 \right) \Big] \\ & = \ ... \end{aligned}
Cheers

too many parentheses within the LaTeX and I just lost it
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Davelittle
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(Original post by Khallil)
I got your back Tenten!



Nope. TenOfThem made a mistake with the simplification by overlooking the parentheses:

\begin{aligned} \sigma & = \dfrac{4}{x} \left( 4\log(x) + 3 \right) - \dfrac{4}{x} \left( 4\log(x) - 3 \right) \\ & = \dfrac{4}{x} \Big[ \left( 4\log(x) + 3 \right) - \left( 4\log(x) - 3 \right) \Big] \\ & = \ ... \end{aligned}
Gotcha, thought the brackets might have been the problem!
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