# Simplify $ln(x(\sqrt{1+e^x}-\sqrt{e^x})) + ln(\sqrt{1+e^{-x}}+1)$

1. Feb 22, 2009

### walker242

1. The problem statement, all variables and given/known data
As in title, simplify $$ln(x(\sqrt{1+e^x}-\sqrt{e^x})) + ln(\sqrt{1+e^{-x}}+1)$$, x > 0.

2. Relevant equations
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3. The attempt at a solution
So far
$$ln(x(\sqrt{1+e^x}-\sqrt{e^x})) + ln(\sqrt{1+e^{-x}}+1) = ln(x(\sqrt{1+e^x}-\sqrt{e^x})(\sqrt{1+e^{-x}}+1)) = ln(x(\sqrt{2+e^{-x} + e^{x}} - \sqrt{e^x}))$$.

Is it possible to get further?

2. Feb 22, 2009

### Staff: Mentor

I worked it through and ended with what you have. I don't see anything more you can do.

3. Feb 22, 2009

Thanks. :)

4. May 8, 2009

### walker242

Actually, it's possible to get further:

$\ln \left(x \left(\sqrt{1+e^{x}}-\sqrt{e^{x}}\right) + \ln\left(\sqrt{1+e^{-x}}+1 \right) = \ln x + \ln\left(\sqrt{1+e^{x}}-\sqrt{e^{x}}\right) + \ln\left(\sqrt{1+e^{-x}}+1\right) =$

$= \ln x + \ln\left(\left(\sqrt{1+e^{x}}-\sqrt{e^x}\right) \left(\sqrt{1+e^{-x}} + 1\right)\right) = \ln x + \ln\left(\sqrt{1+e^{x}}\sqrt{1+e^{-x}}+\sqrt{1+e^{x}}-\sqrt{e^{x}}\sqrt{1+e^{-x}}-\sqrt{e^x}\right)$

$= \ln x + \ln\left(\left(\left(1+e^{x}\right)\left(1+e^{-x}\right)\right)^{\frac{1}{2}} + \left(1+e^{x}\right)^{\frac{1}{2}}-\left(\left(e^{x}\right)\left(1+e^{-x}\right)\right)^{\frac{1}{2}}-\left(e^{x}\right)^{\frac{1}{2}}\right) = \ln x + \ln\left(\left(1+e^{-x}+e^{x}+e^{0}\right)^{\frac{1}{2}} + \left(1+e^{x}\right)^{\frac{1}{2}} - \left(e^{x}+e^{0}\right)^{\frac{1}{2}}-\left(e^{x}\right)^{\frac{1}{2}}\right)$

$= \ln x + \ln\left(\left(2+e^{-x}+e^{x}\right)^{\frac{1}{2}} + \left(1+e^{x}\right)^{\frac{1}{2}} - \left(e^{x}+1\right)^{\frac{1}{2}} -\left(e^{x}\right)^{\frac{1}{2}}\right) = \ln x + \ln\left(\sqrt{\left(2 + e^{-x} + e^{x}\right)} - \sqrt{\left(e^{x}\right)}\right)$

$= \ln x + \ln\left(\sqrt{\left(e^{\frac{x}{2}} + e^{-\frac{x}{2}}\right)^2} - \sqrt{e^x}\right) = \ln x + \ln\left(e^{\frac{x}{2}} + e^{-\frac{x}{2}} - e^\frac{x}{2}\right) = \ln x + \ln\left(e^{-\frac{x}{2}}\right) = \ln \left(\frac{x}{e^\frac{x}{2}}\right)$

5. May 8, 2009

### Staff: Mentor

You're right. It's also possible to get to your result with much less work.
Here's where the OP ended.
$$ln(x(\sqrt{2+e^{-x} + e^{x}} - \sqrt{e^x}))$$
The expression in the first radical happens to be a perfect square trinomial, something I didn't notice earlier.

$$2+e^{-x} + e^{x} = e^{x} + 2 + e^{-x} = (e^{x/2} + e^{-x/2})^2$$
The expression being squared is always positive, so we don't have to worry about the negative square root. IOW,
$$\sqrt{e^{x} + 2+e^{-x} } = \sqrt{(e^{x/2} + e^{-x/2})^2} = e^{x/2} + e^{-x/2}$$

Here's the complete work.
$$ln(x(\sqrt{2+e^{-x} + e^{x}} - \sqrt{e^x})) = ln(x(e^{x/2} + e^{-x/2} - e^{x/2})) = ln(xe^{-x/2}) = ln(\frac{x}{e^{x/2}})$$

Last edited: May 8, 2009