A:

Here we have a quotient inequality of a special nature. We can rearrange it and multiply through by the denominator since the modulus of a function is always positive.

Recall one law of indices .

 or

Solution set

A:

Applying the difference of two squares we have:

Multiplying through by -1 we have

Solution set

A:

Applying the difference of two squares we have:

Multiplying through by -1 we have

Solution set

A:

We may solve this problem by proceeding thus

A:

Recall that

• If
•  or

a) If 

b) If
 or
 or
 or
c) If

A:

Rearranging and factorizing we have

Multiplying the numerator and denominator by , the denominator;
Hence
The zeros are , or

The solution set is

A:

We rearrange and simplify thus

Multiplying the numerator and the denominator by the denominator;

Since the denominator is always positive then the numerator in this problem must also be positive.
Hence
The zeros are  or

The solution set is

A:

Rearranging and factorizing we have

Multiplying the numerator and denominator by , the denominator;
Hence
The zeros are , or

The solution set is

A:

n

We rearrange and simplify as we did in problem 6 above.

 or
Dividing through by -2 , the inequality sign is reversed.

Since the denominator is always positive for all real values of , it implies the numerator must be negative or less than zero

The zeros are  or

The solution set is

A:

Factorizing we have
The zeros are  or

A:

Rearranging and factorising we have

The zeros are , or

The solution set is

A:

Rearranging and factorizing we have

The zeros are  or

Solution set is

A:

Factorizing we have
The zeros are  or

A:

We factorize the inequality thus
Next we find the zeros by putting

Hence we have  or

The solution set is 

A:

If  then

Hence 

A:

Applying change of base we have

A:

Applying the laws of logarithms we have .

The LHS and RHS are logarithms to the same base; hence we compare the functions thus

We divide through by y in order to get an equation in terms of what we need.
Let
 or
Hence  or

A:

We factorize directly in terms of natural logarithms thus
 or

A:

………(1)

……….(2  )

Using equation (1) we change  to base 2 and simplifying the equation we have

Changing from logarithmic to index form we have

……..(3)

From equation (2)
Changing from logarithmic form to index form we have
 …….(4)
Substituting (4) in (3) we have

Factoring out 2 from the bracket we have

Multiplying through by y and dividing through by  we have

Factorizing the quadratic and solving we have

A:

We can prove this by applying logarithms thus

Using the laws of logarithms we have

Grouping like terms we have