Question 326495
Let {{{u=e^x}}}, {{{u^2=e^(2x)}}}
{{{e^(2x)-6e^x+8=0}}}
{{{u^2-6u+8=0}}}
{{{(u-2)(u-4)=0}}}
Two solutions: {{{u=2}}}, {{{u=4}}}
{{{e^x=2}}}
{{{x=ln(2)}}}
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{{{e^x=4}}}
{{{x=ln(4)}}}