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?

Printable View

- Sep 17th 2009, 06:20 PMChinnie15Logarithmic 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 PMskeeter
1) $\displaystyle \ln\left(\frac{1}{\sqrt{1+x^2}} \right) + 1 = 0$

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

change to an exponential equation ...

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

solve for $\displaystyle x$

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

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

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

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

factor the quadratic and solve for $\displaystyle x$ - Sep 17th 2009, 06:37 PMChinnie15
how come you multiply every term by e^x? I don't see that.

- Sep 17th 2009, 06:51 PMskeeter
well, now you've seen it ...

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

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

... and it makes the equation quadratic in $\displaystyle e^x$, which can be easily factored and solved. - Sep 17th 2009, 06:56 PMChinnie15
ok, I see it now. :) thank you!