# © Boardworks Ltd 2014 1 of 7 © Boardworks Ltd 2014 1 of 7 AS-Level Maths: Core 2 for Edexcel C2.6 Exponentials and logarithms This icon indicates the slide.

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© Boardworks Ltd 2014 1 of 7 © Boardworks Ltd 2014 1 of 7 AS-Level Maths: Core 2 for Edexcel C2.6 Exponentials and logarithms This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation.

© Boardworks Ltd 2014 2 of 7 Contents © Boardworks Ltd 2014 2 of 7 Exponential functions Logarithms The laws of logarithms Solving equations using logarithms Examination-style questions Solving equations using logarithms

© Boardworks Ltd 2014 3 of 7 Solving equations involving logarithms We can use the laws of logarithms to solve equations. For example: Solve log 5 x + 2 = log 5 10. To solve this equation we have to write the constant value 2 in logarithmic form: 2 = 2 log 5 5 because log 5 5 = 1 = log 5 5 2 = log 5 25 The equation can now be written as: log 5 x + log 5 25 = log 5 10 log 5 25 x = log 5 10 25 x = 10 x = 0.4

© Boardworks Ltd 2014 4 of 7 Solving equations of the form a x = b We can use logarithms to solve equations of the form a x = b. For example: Find x to 3 significant figures if 5 2 x = 30. We can solve this by taking logs of both sides: log 5 2 x = log 30 2 x log 5 = log 30 Using a calculator: x = 1.06 (to 3 s.f.)

© Boardworks Ltd 2014 5 of 7 Solving equations of the form a x = b Find x to 3 significant figures if 4 3 x +1 = 7 x +2. Taking logs of both sides:

© Boardworks Ltd 2014 6 of 7 Solving equations of the form a x = b Solve 3 2 x –5(3 x ) + 4 = 0 to 3 significant figures. If we let y = 3 x we can write the equation as: So: If 3 x = 1 then x = 0. Now, solving 3 x = 4 by taking logs of both sides:

© Boardworks Ltd 2014 7 of 7 Examination-style question Julia starts a new job on a salary of £15 000 per annum. She is promised that her salary will increase by 4.5% at the end of each year. If she stays in the same job how long will it be before she earns more than double her starting salary? 15 000 × 1.045 n = 30 000 1.045 n = 2 log 1.045 n = log 2 n log 1.045 = log 2 15.7 Julia’s starting salary will have doubled after 16 years.

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