How do I manipulate this to the form desired?

shreddinglicks

1. The problem statement, all variables and given/known data
I want to manipulate an equation to suit a desired form.

2. Relevant equations
$(-h^2/2uc)*(dp/dx)*ln((1+c*(x/h)^2)/(1+c))$

becomes

$-(h^2/2u)*(dp/dx)*(1-(x/h)^2)$
3. The attempt at a solution

I have no idea, I'm not even sure how the natural log disappears.

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Dick

Homework Helper
1. The problem statement, all variables and given/known data
I want to manipulate an equation to suit a desired form.

2. Relevant equations
$(-h^2/2uc)*(dp/dx)*ln((1+c*(x/h)^2)/(1+c))$

becomes

$-(h^2/2u)*(dp/dx)*(1-(x/h)^2)$
3. The attempt at a solution

I have no idea, I'm not even sure how the natural log disappears.
They are using the approximation $\ln(1+a) \approx a$ valid when $a$ is small. Try that along with other rules about logs.

shreddinglicks

They are using the approximation $\ln(1+a) \approx a$ valid when $a$ is small. Try that along with other rules about logs.
I see:

Use $ln(x/y) = ln(x) - ln(y)$

The rest is simple.

Is there a link to somewhere online showing that approximation you gave? Just curious.

Dick

Homework Helper
Is there a link to somewhere online showing that approximation you gave? Just curious.
I don't know any good links. But approximations like this generally come from taking the first terms of the Taylor series. $\ln(1+x)=x- \frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots$ Keeping just the first term gives the approximation. Similarly, $\sin(x) \approx x$ etc.

shreddinglicks

I don't know any good links. But approximations like this generally come from taking the first terms of the Taylor series. $\ln(1+x)=x- \frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots$ Keeping just the first term gives the approximation. Similarly, $\sin(x) \approx x$ etc.
Thanks.

"How do I manipulate this to the form desired?"

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