# Question 1Edit

## iEdit

## iiEdit

## iiiEdit

, then

Either , which means or , which means that .

## ivEdit

## vEdit

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.

## viEdit

Use the same method as **iv**, expand the expression and cancel.

# Question 2Edit

implies that and thus . Step 4 requires division by and thus is an invalid step.

# Question 4Edit

## iiEdit

All values of x satisfy the inequality, since it can be rewritten as , and for all x.

## iiiEdit

If , then .

Thus, , or .

## vEdit

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.

## viEdit

If , then .

Thus , or .

## viiiEdit

Complete the square:

Thus, .

## ixEdit

Solve for first, then consider the third factor for both cases, giving us , or

## xEdit

, or

## xiEdit

If , then taking the base 2 logarithm on both sides:

## xiiEdit

## xiiiEdit

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 5Edit

## iEdit

, which is true since

## iiEdit

## ivEdit

, therefore .

, therefore .

## vEdit

, therefore .

## viiiEdit

If or are 0, then by definition .

Otherwise, we have already proved that for , therefore if , is true.

## ixEdit

If , then since .

If , then . Since , we have .

# Question 6Edit

## iiEdit

If then by **6.i**.

If then and , hence also true.

If then , which means . Therefore, .

## iiiEdit

We already know that if then . If or then or . But since , .

## ivEdit

We already proved in **6.i** that if , then for all .

Thus, if it cannot be or , and must be . If , then , which is positive by virtue of being even.