Solving Logarithmic Equations

Solving Logarithmic Equations

Example 1 Write each equation in exponential form.

3 a. log32 8 = 5

2 The base is 32, and the exponent is 3.

3

8 = 325

1 b. log81 3 = 4

1 The base is 81, and the exponent is 4.

1

3 = 814

Example 2 Write each equation in logarithmic form.

a. 64 = 1296 The base is 6, and the exponent, or logarithm, is 4. log6 1296 = 4

b.

2-8

=

1 256

The base is 2, and the exponent, or

logarithm, is -8.

1 log2 256 = -8

Example 3 1

Evaluate the expression log3 27.

1 Let x = log3 27.

1 x = log3 27

3x

=

1 27

Definition of logarithm

3x = (27)-1

3x = (33)-1 3x = 3-3 x = -3

a-m

=

1 am

33 = 27

(am)n = amn

If au = av, then u = v.

Example 4 CHEMISTRY Refer to the application at the beginning of Lesson 11-4 in your book. How long would it take for 640,000 grams of Polonium-194, with a half-life of 0.5 second, to decay to 5000 grams?

N = N0

1 2

t

5000 = 640,000 1 t 2

1 128 =

1t 2

log

1 2

1 128

= t

log 1

2

1 2 7

= t

log 1

2

17 2

= t

17 = 1t

2

2

7 =t

N

=

N0(1

+

r)t

for

r

=

1 -2

N = 5000, N0 = 640,000

Divide each side by 640,000.

Write the equation in logarithmic form.

128 = 27

1

1n

bn b

Definition of logarithm

It will take 7 half-lives or 3.5 seconds.

Example 5 Solve each equation.

11 a. logp 65614 = 2

11 logp 65614 = 2

1

1

p 2 = 65614

p = 4 6561 ( p)2 = (9)2

p = 81

b. log5 -(5x - 3) = log5 -(10x + 2)

log5 -(5x - 3) = log5 -(10x + 2) -(5x - 3) = -(10x + 2) 5x = -5 x = -1

c. log8 (x + 1) + log8 (x + 3) = log8 24

log8 (x + 1) + log8 (x + 3) = log8 24

log8 [(x + 1)(x + 3)] = log8 24 x2 + 4x + 3 = 24 x2 + 4x - 21 = 0

(x + 7)(x - 3) = 0

x+7 =0

or

x-3 =0

x = -7

x =3

By substituting x = -7 and x = 3 into the equation, we find that x = -7 is undefined for the equation log8 (x + 1) + log8 (x + 3) = log8 24. When x = -7 we get an extraneous solution. So, x = 3 is the correct solution.

Example 6 Graph y = log4 (x + 2).

The equation y = log4 (x + 2) can be written as 4y = x + 2. Choose values for y and then find the corresponding values of x.

y

x + 2

x

(x, y)

-3 0.016 -1.984 (-1.984, -3)

-2 0.063 -1.937 (-1.937, -2)

-1

0.25

-1.75 (-1.75, -1)

0

1

-1

(-1, 0)

1

4

2

(2, 1)

2

16

14

(14, 2)

3

64

62

(62, 3)

Example 7 Graph y log3 x - 4.

The boundary for the inequality y log3 x - 4 can be written as y = log3 x - 4. Rewrite this equation in exponential form.

y

y + 4 3y + 4

= log3 x - 4 = log3 x = x

Use a table of values to graph the boundary.

y

y + 4

x

(x, y)

-7

-3

0.037 (0.037, -7)

-6

-2

0.111 (0.111, -6)

-5

-1

0.333 (0.333, -5)

-4

0

1

(1, -4)

-3

1

3

(3, -3)

-2

2

9

(9, -2)

-1

3

27

(27, -1)

Test a point, for example (0, 0), to determine which region to shade. 3y + 4 0 30 + 4 0 True

Shade the region that contains the point at (0, 0). However, since values of x 0 yield extraneous solutions, only shade above the curve in quadrants I and IV.

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