Example: e to the two x is greater than
e to the x plus twelve.
Setting the right hand side equal to zero,
we get e to the two x minus e to the x,
minus twelve, is greater than zero.
And then using our exponent rules,
we know e to the two x is simply
e to the x squared, minus e to the x, minus
twelve is greater than zero. Now we have a
quadratic, but instead of x's, we have
e to the x. So factoring this quadratic,
we get e to the x minus four, times
e to the x plus three is greater than zero.
So e to the x has to be equal to four or
e to the x has to be equal to negative three,
but we know e to the x cannot equal
negative three, as it has to be greater
than zero. So e to the x equal four is
possible, by taking the natural log of
both sides, we get x is equal to the natural
log of four. Creating our sign chart,
we only have the natural log of four.
Looking at my two factors, we have
e to the x minus four and e to the x plus three,
which we know e to the x plus three is always
positive because e to the x is always positive
and we are adding three. And then
e to the x minus four will be positive to
the right, go through its zero, and then
stay negative. So we have minus and plus
and we are looking for only positive solutions,
so we don't want to include our zero.
So our answer is from the natural log of four,
not included to infinity.