Question 1[edit]










, then

Either
, which means
or
, which means that
.








Note that the two middle terms do no appear to match, but if you follow the logical development in each term, you will see that there is a matching additive inverse for each term on both sides.
Use the same method as iv, expand the expression and cancel.
Question 2[edit]
implies that
and thus
. Step 4 requires division by
and thus is an invalid step.
Question 4[edit]
All values of x satisfy the inequality, since it can be rewritten as
, and
for all x.
If
, then
.
Thus,
, or
.
cannot be factored in its current form, so we first turn a part of the expression into a perfect square:

We then complete the square on
, so now we have the expression:
, which is positive for all values of x.
If
, then
.

Thus
, or
.
Complete the square:

Thus,
.
Solve for
first, then consider the third factor
for both cases, giving us
, or 
, or ![{\displaystyle x<{\sqrt[{3}]{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/61605d902da0e92dfbb747a2a8f8e9a04e2833e9)
If
, then taking the base 2 logarithm on both sides:


NOTE: The answer in the 3rd Edition provides
or
. Plugging in values
, e.g. 10, gives us
, so I think this is a misprint or an incorrect answer.
If
, then
, which is only positive when
since
.
Question 5[edit]


, which is true since 




, therefore
.
, therefore
.
, therefore
.



If
or
are 0, then by definition
.
Otherwise, we have already proved that
for
, therefore if
,
is true.
If
, then
since
.
If
, then
. Since
, we have
.
Question 6[edit]