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Logarithms: Solving equations using logarithms

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1 Logarithms: Solving equations using logarithms
Silent Teacher Intelligent Practice Narration Your Turn 2 π‘₯+1 = 3 π‘₯ 3 2π‘₯ βˆ’2 3 π‘₯ βˆ’15=0 2(7 2π‘₯ )βˆ’ 7 π‘₯+1 +3=0 Example For more videos visit mrbartonmaths.com/videos

2 Worked Example Your Turn Solve the following equations (3sf): Solve the following equations (3sf): 2 π‘₯+1 = 3 π‘₯ 2 π‘₯+1 = 7 π‘₯ 3 2π‘₯ βˆ’2 3 π‘₯ βˆ’15=0 2 2π‘₯ π‘₯ βˆ’18=0 @mrbartonmaths

3 Solve the following equations for the value of π‘₯ (3sf):
5 π‘₯ = 2 π‘₯+1 7 π‘₯+1 = 3 π‘₯+2 2 3π‘₯ = 5 (2π‘₯+1) 7 π‘₯+3 = 5 π‘₯+4 3 2π‘₯ βˆ’15 3 π‘₯ +44=0 5 2π‘₯ βˆ’ 5 π‘₯ βˆ’2=0 2( 3 2π‘₯ )βˆ’7 3 π‘₯ βˆ’4=0 3 2π‘₯ π‘₯ βˆ’2=0 2(7 2π‘₯ )βˆ’ 7 π‘₯+1 +3=0 @mrbartonmaths

4 Solve the following equations for the value of π‘₯ (3sf):
5 π‘₯ = 2 π‘₯+1 0.757 7 π‘₯+1 = 3 π‘₯+2 0.297 2 3π‘₯ = 5 (2π‘₯+1) βˆ’1.41 7 π‘₯+3 = 5 π‘₯+4 1.78 3 2π‘₯ βˆ’15 3 π‘₯ +44=0 1.26, 2.18 5 2π‘₯ βˆ’ 5 π‘₯ βˆ’2=0 0.431 1.26 2( 3 2π‘₯ )βˆ’7 3 π‘₯ βˆ’4=0 βˆ’1 3 2π‘₯ π‘₯ βˆ’2=0 2(7 2π‘₯ )βˆ’ 7 π‘₯+1 +3=0 βˆ’0.356, 0.565 @mrbartonmaths


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