Exercise 2 Page 488

We can use the fact that log_4 3 ≈ 0.7925 and log_4 5 ≈ 1.1610 to approximate the value of the given expression. To do so, we will first recall three useful Properties of Logarithms.

	Properties of Logarithms
Name	Product Property	Quotient Property	Power Property
Condition	a>0, b>0, and x ≠ 1	a>0, b>0, and x ≠ 1	m>0, b>0, and b ≠ 1
Property	log_x ab = log_x a + log_x b	log_x ab=log_x a - log_x b	log_b m^p = p log_b m

Let's now find the approximation of log_4 15. We will start by writing 15 as a product of factors. log_4 15 ⇔ log_4 (5* 3) Now, we will use Product Property of Logarithms to use the given values of logarithms. log_4 (5* 3) ⇔ log_4 5 + log_4 3 Finally, let's substitute 0.7925 for log_4 3 and 1.1610 for log_4 5.

log_4 5 + log_4 3

SubstituteII

log_4 3= 0.7925, log_4 5= 1.1610

0.7925 + 1.1610

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1.9535