## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 7 - Exponential Functions - 7.3 Logarithms and Their Derivatives - Exercises - Page 342: 14

#### Answer

$$5$$

#### Work Step by Step

$\log_2 {\frac{4}{3}}+\log_2 {24}$ $=\log_2 ({\frac{4}{3}\times {24}})$ {Since $\log{a}+\log{b}=\log{ab}$} $=\log_2{32}=\log_2{2^5}=5$

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