11 ALGEBRA

Since 3a – 2b = 5 and 3b – 2a = 5, we can write

3a – 2b = 3b – 2a 3a – 3b + 2a – 2b = 0.

The only possible solution to this equation is a = b, since both a > b and a < b will give a result different from zero. Therefore, we have

3a – 2a = 5.

Obviously, a = 2 is a solution. We can also write this equation as

which leads us to conclude that 3a – 2a = 5 has a unique solution (can you see why?). Thus a = b = 2 and

log2(x – 3) = a = 2 x – 3 = 22 = 4 x = 7 .

Check Yourself 33

Solve the equations.

f. Logarithmic Equations with Parameters

Sometimes we may be asked to solve an equation for the values of a parameter which satisfy a specific condition, or to investigate the solution(s) of an equation for different values of a parameter.

Substituting loga x = y gives us

2. Logarithmic Inequalities

Sometimes we are asked to solve inequalities between two functions of the form logf(x)g(x)

and logf(x)h(x), for example: logf(x)g(x) > logf(x)h(x), logf(x)g(x) logf(x)h(x), etc. In this case we can establish the systems

We can solve the resulting systems using a sign table. The values satisfying any of these systems will be included in the solution set of the original inequality.

For logarithmic inequalities which do not have functions of the form logf(x) g(x) and logf(x) h(x), we first apply the properties of logarithms in order to write them in this form.

Let us look at some examples.

Solution a. Using the monotone property of logarithmic functions and the existence conditions for

b. Since the base is between zero and 1, we establish the system

c.The right-hand side of the inequality is not in logarithmic form. However, we can use the identity log2 2 = 1 and write

By considering the critical values, we establish the sign table:

Therefore the solution is x (– , –9] (3, ).

Note

For inequalities of the form loga g(x) > b or logag(x) < b, we can use the property b = loga ab and write

for a > 1: loga g(x) b 0 g(x)b ab loga g(x) b g(x) a ,

loga g(x) b g(x) ab loga g(x) b 0 g(x) ab.

e.We have log(x2 – 5x + 7) 0 log(x2 – 5x + 7) log 1. Establishing the appropriate system, we get

f. By the properties of logarithms, we can write the inequality as log2x(5x – 3) 1 log2x(5x – 3) log2x(2x).

loga a =1

There are two cases: the base may be between zero and 1, or it may be greater than 1. Accordingly we have the two systems

which leads us to the solution y (– , –2) (–ñ2, –1) (0, ñ2).

We can find the corresponding values of x by considering the three cases: For y (– , –2), y = log2 x –2 0 x 2–2 0 x 14 .

For y (0, ñ2), 0 log2 x ñ2 1 x 2ñ2.

Therefore the solution is x (0, 14 ) ( 212 , 21 ) (1, 2ñ2).

C. SYSTEMS OF EQUATIONS AND INEQUALITIES

1. Systems of Equations

We can solve systems of exponential or logarithmic equations by using the properties we have studied so far combined with algebraic operations, elimination and substitution. We write the solution of a system as a set of ordered pairs, triples or quadruples, etc. depending on the number of unknowns in the system.

Hence {(3, 2)} is the solution.

EXAMPLE 103 Solve the systems of equations.

To satisfy the existence conditions, x and y must be positive. Therefore we eliminate the case x = –5 and write the solution as {(3, 5)}.

2. Systems of Inequalities

To solve a system of exponential or logarithmic inequalities, we consider each inequality separately and find its solution. Then we add the existence conditions of logarithms to the system and find a common solution set. We can also use a sign table to look at the different cases.

The following examples illustrate this approach.

EXAMPLE 104 Solve the systems of inequalities.