Solving for x in a equation involving natural logarithms

Question

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.

Use the fact that $\log (xy) = \log x + \log y$. Then exponentiate. — copper.hat, Oct 09 '13 at 20:19

score 3 · Accepted Answer · answered Oct 09 '13 at 20:20

3

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

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

answered Oct 09 '13 at 20:20

what'sup

score 2 · Answer 2 · answered Oct 09 '13 at 20:36

2

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

answered Oct 09 '13 at 20:36

1190

score 0 · Answer 3 · edited Oct 09 '13 at 20:40

0

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

edited Oct 09 '13 at 20:40

Davide Giraudo

answered Oct 09 '13 at 20:22

drhab

score 0 · Answer 4 · answered Oct 09 '13 at 20:27

0

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

answered Oct 09 '13 at 20:27

Albert Zhang

4 Answers4