Question 785014
If u = {{{e^x}}} then {{{u^2 = e^(2x)}}} - when multiplying a base to a power by the same base to a different power, add the powers.  {{{e^x * e^x = e^(x+x) = e^(2x)}}}<P>
{{{e^2x - e^x =6 }}} means {{{u^2 -u=6}}}<P>
{{{u^2 - u - 6 = 0}}}<P>
(u-3)(u+2) = 0<P>
u = 3 and u = -2<P>
{{{e^x = 3}}} Use the inverse function of e which is ln.<P>
{{{ln(e^x) = ln(3)}}}<P>
x = ln(3) = approximately 1.0986<P>
For u=-2, do the same.  {{{e^x = -2}}}<P>
{{{ln(e^x) = ln(-2)}}}<P>
x = ln(-2):  ln of a negative number is not a real number.  So discard this answer.  If you want to find the imaginary number solution, it's {{{ln(2) + pi*i}}} = approximately {{{.69315 + pi*i}}}<P>
{{{e^(-2x) - 3e^(-x) = -2}}} is {{{1/u^2 -3/u = -2}}}<P>
Multiply all terms by {{{u^2}}} to get {{{1 - 3u = -2u^2}}}<P>
{{{2u^2 -3u + 1 = 0}}}<P>
(2u-1)(u-1) = 0<P>
2u = 1 so u = 1/2 so {{{e^x = 1/2}}} and thus x = ln(1/2) = approximately -.693<P>
u=1 so {{{e^x = 1}}} so x = ln(1) = 0