Given that ln a = 5 , express as a simple natural logarithm: 1 2 log a x

Given, 1 2 log a x From the change of base formula, we know that log b a = log c a log c b where a , b , c > 0 and b ≠ 1 , c ≠ 1 Thus, 1 2 log a x = 1 2 log e x log e a = 1 2 ln x ln a Hence, log a x 2 = 1 2 ln x 5 = 1 10 ln x ∵ ln a = 5

Given that ln a = 1 , express as a single logarithm: 3 log a x + 1 2 log a x

Given, 3 log a x + 1 2 log a x and ln a = 1 Consider 3 log a x + 1 2 log a x = 7 2 log a x = 7 2 log e x log e a = 7 2 ln x ln a = 7 2 ln x ∵ ln a = 1 Hence, 3 log a x + 1 2 log a x = 7 2 ln x

Given that ln a = 1 , express as a single logarithm: 2 ln x 3 - 3 2 log a x

Given, 2 ln x 3 - 3 2 log a x and ln a = 1 Consider 2 ln x 3 - 3 2 log a x = 2 ln x 3 - 3 2 log e x log e a [Using change of base formula] = 2 ln x 3 - 3 2 ln x ln a = 2 ln x 3 - 3 2 ln x ∵ ln a = 1 = 2 ln x - 2 ln 3 - 3 2 ln x = 1 2 ln x - 2 ln 3 = ln x - ln 3 2 ∵ ln p q = q ln p , where p > 0 , q ∈ R = ln x - ln 9 = ln x 9 ∵ ln p q = ln p - ln q , where p , q > 0 Hence, 2 ln x 3 - 3 2 log a x = ln x 9

E M B I B E

Extended MYP Mathematics A Concept based approach Years 4 & 5>Go Ahead and Log In>Chapter 16 - Laws of Logarithms>Practice 5

Rose Harrison, Clara Huizink, Aidan Sproat Clements and, Marlene Torres Skoumal Solutions for Exercise 14: Practice 5

Author:Rose Harrison, Clara Huizink, Aidan Sproat Clements & Marlene Torres Skoumal

Rose Harrison Mathematics Solutions for Exercise - Rose Harrison, Clara Huizink, Aidan Sproat Clements and, Marlene Torres Skoumal Solutions for Exercise 14: Practice 5

Attempt the free practice questions from Exercise 14: Practice 5 with hints and solutions to strengthen your understanding. Extended MYP Mathematics A Concept based approach Years 4 & 5 solutions are prepared by Experienced Embibe Experts.