# Logarithmic Equations

• Sep 17th 2009, 06:20 PM
Chinnie15
Logarithmic Equations
1) ln(1/sqrt(1+x^2)) +1=0
???
2) e^x=1+2e^-x

for this one I've got e^(x)-2e=1, then e(x-2)=1. I'm just unsure of where to go from here?
• Sep 17th 2009, 06:30 PM
skeeter
Quote:

Originally Posted by Chinnie15
1) ln(1/sqrt(1+x^2)) +1=0
???
2) e^x=1+2e^-x

for this one I've got e^(x)-2e=1, then e(x-2)=1. I'm just unsure of where to go from here?

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

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

change to an exponential equation ...

$e^{-1} = \frac{1}{\sqrt{1+x^2}}$

solve for $x$

2) $e^x = 1 + 2e^{-x}$

multiply every term by $e^x$ ...

$e^{2x} = e^x + 2$

$e^{2x} - e^x - 2 = 0$

factor the quadratic and solve for $x$
• Sep 17th 2009, 06:37 PM
Chinnie15
how come you multiply every term by e^x? I don't see that.
• Sep 17th 2009, 06:51 PM
skeeter
Quote:

Originally Posted by Chinnie15
how come you multiply every term by e^x? I don't see that.

well, now you've seen it ...

I wanted to get rid of the $e^{-x}$ in the equation.

it's like if you had the equation $y = 1 + \frac{2}{y}$ ... you'd multiply every term by $y$ to clear the unwanted fraction.

... and it makes the equation quadratic in $e^x$, which can be easily factored and solved.
• Sep 17th 2009, 06:56 PM
Chinnie15
ok, I see it now. :) thank you!