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How would I solve for x in this equation here:

$$\ln(x)+\ln(1/x+1)=3$$

I realize that the answer is $e^3-1$, but I am not sure as to how to get it. Any input is appreciated.

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  • $\begingroup$ Use the fact that $\log (xy) = \log x + \log y$. Then exponentiate. $\endgroup$
    – copper.hat
    Oct 9, 2013 at 20:19
  • $\begingroup$ Thanks, how silly of me to forget that. $\endgroup$
    – Iceandele
    Oct 9, 2013 at 20:31

4 Answers 4

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$$ \ln x + \ln \left( \frac{1}{x} + 1 \right) = \ln \left( x(1/x + 1) \right) = 3 $$

$$ \Rightarrow \ln(x+1)= 3 \Rightarrow x + 1 = e^3 \Rightarrow x = e^3 - 1 $$

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We know that $ln(a)+ ln(b) =ln(ab)$

Then,

$$ln(x)+ln(\frac{1}{x})=ln(x(\frac{1}{x}+1)=ln(1+x)=3$$

Now,

We know that $e^{ln(a)}=a$

Then, $$e^{ln(1+x)}=e^3 $$

This implies that $$x+1=e^3 \implies x=e^3-1$$

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Use that $\ln{x}+\ln{y}=\ln{xy}$ to simplify; the equation becomes much simpler.

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Hint: $\ln(x)+\ln\left(\frac 1{x+1}\right)=\ln\left(\frac x{x+1}\right)$.

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