11 ALGEBRA

Replacing these values in (1) gives us (m + 6) (m – 2) > 0

–m + 3 < 0,

from which we conclude m > 3. Similarly, (2) becomes (m + 6) (m – 2) = 0

m > 0,

which gives us m = 2. Therefore, 4x – (m 2x) – m + 3 = 0 has only one solution if m {2} (3, ).

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1. Determine all values of a for which the equation 9x + a (a + 1) = (2a + 1) 3x has a. no solution. b. only one solution. c. two solutions.

2.State the solutions of the equation 4x – 2m(m + 1) 2x – 1 + m3 = 0 in terms of the parameter m.

2. Exponential Inequalities

To solve simple exponential inequalities, we use the monotone property of exponential functions which we saw earlier in this module:

Solution a. We can write 24x < 16 24x < 24. Since the base (2) is greater than 1, the direction of inequality for the exponents will be the same: 4x < 4 x < 1.

So the inequality holds for x (– , 1).

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1.Solve (x2 + x + 1)x < 1.

2.Solve 4x – a 2x + 3 – a 0 for different real values of the parameter a.

B. LOGARITHMIC EQUATIONS AND INEQUALITIES

1. Logarithmic Equations

An equation containing logarithms which have the variable in their base or argument (or both) is called a logarithmic equation. Here are some examples of logarithmic equations:

log2 x = 3, log2(x x+2)= 2, logx 3 + logx 9 = –3, logx(2x + 3) = 2. When solving logarithmic equations, we can use the following properties of logarithms:

1.loga(f(x) g(x)) = loga|f(x)| + loga|g(x)|

2.loga( gf((xx))) = loga|f(x)| – loga|g(x)|

3.loga f n(x) = n loga|f(x)|.

We must also remember that the argument of a logarithm is always positive, and its base is always positive and different from 1. As we have seen, these conditions are called the existence conditions for logarithms.

Just as we saw for exponential equations, there is no single method that we can use to solve all logarithmic equations. Instead, we can consider some common types of equation and their solution. As a general rule, after solving a given equation we always need to check which solutions satisfy the existence conditions in order to complete the solution.

a. Equations of the Form logf(x) g(x) = b

To solve an equation of this type, we can write it in exponential form using logf(x) g(x) = b f(x)b = g(x) and solve this new equation for x.

After finding the solution(s), we eliminate any values of x which do not satisfy the restrictions f(x) 0, f(x) 1, and g(x) 0 (by the existence conditions).

Let us look at some examples.

By the existence conditions, since x = 3 satisfies 3x – 1 > 0 it is the solution of the given logarithmic equation.

b.logx+1 16 = 2 (x + 1)2 = 16 x + 1 = 4 x1 = 3 or x2 = –5

Check against the existence conditions: we need x + 1 > 0 and x + 1 1, so we accept only x = 3 as the solution since x = –5 does not satisfy x + 1 > 0.

c.log3(5x2 – 9x + 7) = 2 5x2 – 9x + 7 = 32 5x2 – 9x – 2 = 0 (5x + 1) (x – 2) = 0

Accordingly, we get x1 = – 51 and x2 = 2 as possible solutions. Checking for 5x2 – 9x + 7 > 0, we decide that both solutions satisfy the inequality. So the solution set is S ={– 51, 2}.

(x + 4) (x – 1) = 0 x1 = –4 and x2 = 1

If we check these solutions for the conditions 2x > 0, 2x 1 and 3x2 – 3x + 4 > 0, we find that x1 = –4 does not satisfy 2x > 0. Therefore we eliminate it, and so the only solution is x = 1 .

When x = 5, 5x = 25 0 and log5 ò5x = log5 ó5 5 = log5 5 = 1 0 and so we conclude that x = 5 is the solution of the equation.

c. log2(3 – x) + log2(1 – x) = 3 log2[(3 – x) (1 – x)] = 3 x2 – 4x + 3 = 23 x2 – 4x – 5 = 0 (x – 5) (x + 1) = 0 x1 = –1 and x2 = 5

Checking for the conditions 3 – x > 0 and 1 – x > 0, we conclude that x = 5 does not satisfy either condition. Therefore we eliminate it. Since x = –1 satisfies both conditions, it is the only solution to the given equation.

Check Yourself 29

Solve the equations.

b. Equations of the Form logf(x) g(x) = logf(x) h(x)

Since logarithmic functions are one-to-one, we can write logf(x) g(x) = logf(x) h(x) g(x) = h(x).

Therefore, to solve an equation of the form logf(x)g(x) = logf(x)h(x), we solve the equation g(x) = h(x) for x and then check the solutions for the restrictions f(x) 0, f(x) 1, g(x) 0 and h(x) 0.

EXAMPLE 94 Solve the equations.

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Solve the equations.

c. Working with Logarithmic Equations

Both values of x satisfy the condition 12 – 2x >0, so the solution set is S = {2, 3}.

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Solve the equations.

a. log2(9 – 2x) = 3 – x b. xlog3 3x = 9 c. x2+log 2 x = 256 d. log2(9x–1 + 7) = 2 + log2(3x–1 + 1)

d. Equations with Logarithms to Different Bases

In order to work easily with logarithms, we need to have them to the same base. We can use the following properties of logarithms to help us:

For a, b, c 0 and a, c 1,

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e. Logarithmic Equations with a Unique Solution

If we can identify that a given equation has a unique solution (for example: by using the monotony of functions, or by using an arithmetic or geometric argument), and if we observe that a particular value of the variable satisfies the equation, then we do not need to look for another solution.

Solution a. We can observe that x = 3 satisfies the equation, since log3 3 = 4 – 3 = 1.

In addition, since the left-hand side of the given equation is strictly increasing and the righthand side is strictly decreasing, we can conclude that this solution is unique: x = 3 .

b.Let log2(x – 3) = a and log3(x + 2) = b. So 3a – 2b = 5. In addition, we can write

log2(x – 3) = a 2a = x – 3 2a + 3 = x and log3(x + 2) = b 3b = x + 2 3b – 2 = x,

which gives us 2a + 3 = 3b – 2, i.e. 3b – 2a = 5.