e^2x-7e^x+10
bit confused on how to find the x?
$\displaystyle e^{2x}-7e^x+10=0$
$\displaystyle (e^{x})^2-7e^x+10=0$
Lets make $\displaystyle e^{x} = a$
Now we have
$\displaystyle a^2-7a+10=0$
which can be factored as follows
$\displaystyle (a-5)(a-2)=0$
by the null factor law
$\displaystyle a=2,5$
and bring our original unknown back into the picture
$\displaystyle e^x=2,5$
$\displaystyle x=\ln(2),\ln(5)$
Does this make sense?