How would I go about doing problems such as
e^x+1=1/e
and
e^x+4=1/e^2x
Thanks!
$\displaystyle e^x+1=\frac{1}{e}$
$\displaystyle e^x=\frac{1}{e}-1$
$\displaystyle x=\ln\left(\frac{1}{e}-1\right)$
$\displaystyle e^x+4=\frac{1}{e^{2x}}$
$\displaystyle e^{2x}(e^x+4)=1$
$\displaystyle e^{2x}(e^x+4)=1$
$\displaystyle e^{3x}+4e^{2x}=1$
Now substituting
$\displaystyle a = e^x$
$\displaystyle a^{3}+4a^{2}-1=0$
You now have to solve this cubic.
You should consider this
$\displaystyle e^x = y \Rightarrow x = \log_e(y)$
or
$\displaystyle e^x = y \Rightarrow x = \ln(y)$
Therefore using the second transformation
$\displaystyle e^{x+1}=\frac{1}{e} $
$\displaystyle x+1=\ln\left(\frac{1}{e}\right) $
$\displaystyle x=\ln\left(\frac{1}{e}\right)-1 $