Chapter 5 Logarithmic Functions.

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Presentation transcript:

Chapter 5 Logarithmic Functions

5.4 Properties of Logarithms

Exponential/Logarithmic Forms Property For a > 0, b > 0, and b ≠ 1, the equations logb(a) = c and bc = a are equivalent.

Example: Solving Logarithmic Equations in One Variable Solve for x. 1. 6 log9(t) + 1 = 4 2. log3(x4) = 2

Solution 1. Get log9(t) alone on the left side of the equation and solve for t:

Solution 2. Write log3(x4) = 2 in exponential form and solve for x:

Example: Solving for the Base of a Logarithm Solve for b. 1. logb(81) = 4 2. logb(67) = 5

Solution 1. 2.

Summary For an equation of the form logb(x) = k, we can solve for b or x by writing the equation in exponential form.

Power Property for Logarithms For x > 0, b > 0, and b ≠ 1, logb(xp) = p logb(x) In words, a logarithm of a power of x is the exponent times the logarithm of x.

Logarithm Property of Equality For positive real numbers a, b, and c, where b ≠ 1, the equations a = c and logb(a) = logb(c) are equivalent.

Example: Solving an Exponential Equation Solve the equation 2x = 12.

Solution Check: 23.5850 ≈ 12.0003 ≈ 12

Computing a Quotient with a Calculator Warning To compute the quotient in the previous example:

Example: Solving an Exponential Equation Solve 3(4)x = 71.

Solution Check: 3(4)2.2824 ≈ 71.0008 ≈ 71

Dividing Both Sides of an Equation Warning To solve some equations of the form abx = c for x, we divide both sides of the equation by a, and then take the log of both sides. Next, use the power property for logarithms. Note that in the previous example, 3(4)x ≠ (3∙4)x

Example: Solving an Exponential Equation Solve 7(2)x – 4 = 20 + 3(2)x.

Solution

Solution To check on a graphing calculator, graph the equations y = 7(2)x – 4 and y = 20 + 3(2)x and find the approximate intersection point (2.5850, 38), which has x-coordinate 2.5850. This checks.