11 ALGEBRA

24.The San Francisco earthquake of 1989 measured

7.1on the Richter scale. Compare this with the earthquake in Indonesia in 1985 which measured

25.If one earthquake is 30 times more intense than another, how much larger is its measurement on the Richter scale?

26.The Alaskan earthquake of 1964 had a magnitude of 8.6 on the Richter scale. How many times more intense was this than the 1999 Marmara earthquake which measured 7.4 on the scale?

27.Use the Richter scale formula to find the magnitude of an earthquake with the given intensity.

a. 100 times the intensity of I0

b. 10,000 times the intensity of I0 c. 100,000 times the intensity of I0

28.What is the pH of a solution whose concentration of hydrogen ions is 6.5 10–8 moles per liter?

29.The most acidic rainfall ever measured occurred in Scotland in 1974. Its pH was 2.4. Find the hydrogen ion concentration of this rain.

30.Most solutions have a pH value between 1 and 14. Find the corresponding range of [H+].

31.Find the loudness of a dishwasher that operates at an intensity of 10–5 W/m2.

32.The intensity of a sound A is 50 times the intensity of a sound B. What is the difference between the two sound levels in decibels?

33.A man hears a noise which has an intensity of 10–9.8. What is the sound level of the noise in decibels?

Mixed Problems

34.f(x) = ñx + 5, g(x) = log3(5x + 1) and (f g–1)(4) = m2 are given. What is m?

35.Compare a and b in each case.

11

a. a = log 1 4, b = log 13 53

b. a = log4 3, b = log5 3

37. Find a if f(x) = log3(4x + a) and f –1(4) = 2.

If f(x) is one-to-one then f(x) = f(y) x = y.

For f(x) = ax,

ax = ay x = y.

EXAMPLE 76

Solution

y is the multiplicative inverse of x if x y = 1.

( ab )m =( ab )– m

1. Exponential Equations

An exponential equation is an equation in which the unknown appears only in the exponent(s). For example, 3x+2 = 1 is an exponential equation, but x 3x = 3 is not an exponential equation. A solution of an exponential equation is a value of x which satisfies the equation. The set which contains all the solutions of an exponential equation is called the solution set of the equation. Two exponential equations are said to be equivalent if they have exactly the same solution set. Solving an exponential equation means finding its solution set.

There is no single method that we can use to solve all exponential equations. However, we can apply some principles to solve certain types of equation. Let us look at these different types in turn.

a. Equations of the Form af(x) = ag(x)

Since exponential functions are one-to-one, af(x) = ag(x) means f(x) = g(x). So the equations f(x) = g(x) and af(x) = ag(x) are equivalent: they have the same solution set.

Solve each equation for x.

a. 23x = 29 b. 7x+9 = 72x – 15

a.The expressions on each side of the equation have the same base (2). So we can simply equate the exponents:

23x = 29 3x = 9 x = 3.

b.Since the bases are the same, the exponents must be equal. So we can write x + 9 = 2x – 15, which gives us x = 24.

c.First we need to make the bases equal. Since 113 is the multiplicative inverse of 113 ,

(113 )3x–10 =(113 )7 x–10 (113 )3 x–10 =(113 )–(7 x–10) (113 )3 x–10 =(113 )10 – 7 x.

Now we can solve the equation by equating the exponents: 3x – 10 = 10 – 7x 10x = 20 x = 2.

EXAMPLE 77

Solution

loga a = 1

loga x+loga y = loga (x y)

We can keep the solution in this form or we can write it as a single logarithm:

c. Since –1 < 0, the equation will have no solution. So the solution is the empty set.

EXAMPLE 78

Solution

loga b = logc b logc a

Check Yourself 16

Solve the equations.

c. Equations of the Form af(x) = bg(x)

If an equation has the form af(x) = bg(x) and the bases a and b cannot be equalized easily, we can take the logarithms of both sides to a convenient base (usually base 10) and then solve the new equation.

a. Since the bases are different, we take the common logarithms of both sides: log (2x+1) = log (3x – 2) (x + 1) log 2 = (x – 2) log 3

(x log 2) + log 2 = (x log 3) – 2 log 3 log 2 + (2 log 3) = (x log 3) – (x log 2)

EXAMPLE 79

Solution

an > 0

b. Taking the common logarithms of each side, we get

2x – 1 5x+3 = 3 log (2x – 1 5x+3) = log 3 log 2x – 1 + log 5x+3 = log 3 (x – 1) log 2 + (x + 3) log 5 = log 3

(x log 2) – log 2 + (x log 5) + 3 log 5 = log 3

Check Yourself 17

Solve the equations.

d. Working with Exponential Equations

Not all exponential equations fall into the three categories we have seen. If an equation is in a different form, we use methods such as factorization, substitution and division to obtain an equivalent equation which we can solve.

a.By rewriting the given equation as (3x)2 – 3x = 72 and using the substitution y = 3x, we get the quadratic equation

y2 – y = 72 y2 – y – 72 = 0. We can solve this by factorization:

(y – 9) (y + 8) = 0 y1 = 9 and y2 = –8.

Now we must be careful. Since y = 3x is always positive, we must reject the negative solution y2 = –8. Only y = 9 will be used to solve the original equation:

3x = y = 9 3x = 32 x = 2. So the solution set is S = {2}.

b. First we rewrite the equation:

4x – 2x+4 + 48 = 0 22x – (24 2x) + 48 = 0 (2x)2 – (16 2x) + 48 = 0. Now we can substitute y = 2x and get a quadratic equation:

y2 – 16y + 48 = 0 (y – 4) (y – 12) = 0 y1 = 4 and y2 = 12. Back substituting both y = 4 and y = 12, we get

2x = y = 4 2x = 22 x = 2 and 2x = y = 12 x = log2 12. So the solution set is S = {2, log2 12}.

EXAMPLE 80

Solution

Check Yourself 18

Solve the equations.

Factorizing this as (y – 1) (y – 3) = 0 gives us the solutions y1 = 1 and y2 = 3. Since both solutions are positive, we need to solve both 33x = 1 and 33x = 3:

33x = 1 3x = 0 x = 0 and 33x = 3 3x = 1 x = 13. So the solution set is S ={0, 13}.

EXAMPLE 81

Solution

Check Yourself 19

Solve the equations.

a. 2x+1 – 22–x = 7 b. (2 + ñ3)x + (2 – ñ3)x = 14

a. 2x+4 + 2x+2 = 5x+1 + (3 5x) 2x 24) + (2x 22) = (5x 5) + (3 5x)

2x (16 + 4) = 5x (5 + 3) 2x = 8 ( 2 )x = 2 x =1 5x 20 5 5

b. We begin by rearranging the equation so that similar bases are grouped on the same side: