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

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In summary, the given expression can be simplified to ln(\frac{x}{e^{x/2}}). It is also possible to get to this result by recognizing that the expression in the first radical is a perfect square trinomial, simplifying to e^{x/2} + e^{-x/2}.
  • #1
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Homework Statement


As in title, simplify [tex]ln(x(\sqrt{1+e^x}-\sqrt{e^x})) + ln(\sqrt{1+e^{-x}}+1)[/tex], x > 0.


Homework Equations


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The Attempt at a Solution


So far
[tex]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}))[/tex].

Is it possible to get further?
 
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  • #2
I worked it through and ended with what you have. I don't see anything more you can do.
 
  • #3
Thanks. :)
 
  • #4
Actually, it's possible to get further:

[itex]\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) =[/itex]

[itex]= \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)[/itex]

[itex]= \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)[/itex]

[itex]= \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)[/itex]

[itex]= \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)[/itex]
 
  • #5
You're right. It's also possible to get to your result with much less work.
Here's where the OP ended.
[tex] ln(x(\sqrt{2+e^{-x} + e^{x}} - \sqrt{e^x}))[/tex]
The expression in the first radical happens to be a perfect square trinomial, something I didn't notice earlier.


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

Here's the complete work.
[tex]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}})[/tex]
 
Last edited:

What is the purpose of simplifying $ln(x(\sqrt{1+e^x}-\sqrt{e^x})) + ln(\sqrt{1+e^{-x}}+1)$?

The purpose of simplifying this expression is to make it easier to work with and to potentially find patterns or relationships that can be used in other mathematical calculations.

Can this expression be simplified further?

Yes, this expression can be simplified further by using logarithm rules and properties.

What are the common mistakes made when simplifying this expression?

Some common mistakes include not distributing the logarithm or not recognizing logarithmic properties such as the product and quotient rules.

What is the final simplified form of this expression?

The final simplified form of this expression is $ln(x) + ln(\sqrt{1+e^{-x}}+1)$.

How can this simplified expression be used in other mathematical calculations?

This simplified expression can be used in other mathematical calculations by substituting in specific values for x to solve for the natural logarithm of a number. It can also be used to simplify other expressions involving logarithms.

Suggested for: Simplify $ln(x(\sqrt{1+e^x}-\sqrt{e^x})) + ln(\sqrt{1+e^{-x}}+1)$

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