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Page 1: Indices, Surds and Logarithms Notes - Mulberry Education - Indices, Surds and Logarithms...Microsoft Word - Indices, Surds and Logarithms Notes.docx Keywords Created Date: 20151008203621Z

 

263  Tanjong  Katong  Rd  #01-­‐07,  Tel:  67020118  

   

Indices,  Surds  and  Logarithms  

Logarithms  Common  Laws  

log! 𝑏 + log! 𝑐 = log!(𝑏𝑐)  

log! 𝑏 βˆ’ log! 𝑐 = log!𝑏𝑐  

log! 𝑏! = 𝑐 log! 𝑏                                                                            log! 𝑐 =

!"#! !!"#! !

 (b  can  take  any  value)    Converting  from  log  from  to  index  Form  

log! 𝑏 = 𝑐       ↔    π‘ = π‘Ž!  Example:  

log! 8 = 3     ↔    8 = 2!    Common  &  Natural  logarithms  

log!" π‘Ž = lg π‘Ž  log! π‘Ž = ln π‘Ž  

(Use  β€œln”  when  there  is  π‘’  ;  otherwise  always  use  β€œlg”)    Solving  Equations  2  Scenarios  to  reach:  1.  A  single  log  on  both  sides  with  the  same  base  Example:  

log!(2π‘₯ + 1) = log!(π‘₯ βˆ’ 3)  2π‘₯ + 1 = π‘₯ βˆ’ 3  

π‘₯ =βˆ’3 βˆ’ 12

= βˆ’2    2.  A  single  log  on  one  side  of  the  equation  Example:  

log! 2π‘₯ + 1 = 2  2π‘₯ + 1 = 3!  

π‘₯ =9 βˆ’ 12

= 4  

 

Surds  Common  Laws  

π‘Ž  Γ— 𝑏 = π‘Žπ‘  π‘Žπ‘=

π‘Žπ‘  

π‘Ž  Γ— π‘Ž = π‘Ž  π‘Ž 𝑏  Γ—𝑐 𝑑 = π‘Žπ‘ 𝑏𝑑  

π‘Ž + 𝑏 𝑐 + 𝑑 = π‘Žπ‘ + π‘Ž 𝑑 + 𝑐 𝑏 + 𝑏𝑑  1 + π‘Ž 1 βˆ’ π‘Ž = 1! βˆ’ π‘Ž

!= 1 βˆ’ π‘Ž  

 Rationalising  the  denominator  To  remove  the  surd  from  the  denominator,  multiply  the  entire  fraction  by  the  conjugate  surd      Example:    

34 βˆ’ 2 6

=3

4 βˆ’ 2 6Γ—4 + 2 64 + 2 6

 

=12 + 6 6

4! βˆ’ 2 6!  

=6(2 + 6)16 βˆ’ 4(6)

 

= βˆ’3(2 + 6)

4  

 Simplfying  surds  Split  the  surd  into  a  square  root  of  a  perfect  square  times  another  surd:    Examples:    

200 = 100    Γ— 2 = 10 2    768 = 256  Γ— 3 = 16 3  125 = 25  Γ— 5 = 5 5  

Indices  β€’ π‘Ž!Γ—π‘Ž! = π‘Ž!!!  β€’ π‘Ž! Γ· π‘Ž! = π‘Ž!!!  β€’ π‘Ž!! = !

!!  

β€’ !!

!= !!

!!  

β€’ !!

!!= !

!

!= !!

!!  

β€’ !!!!

!!!!= !!!

!!!  

β€’ π‘Ž! = 1  β€’ π‘Ž

!! = π‘Ž  

β€’ π‘Ž!! = π‘Ž!  

β€’ π‘Ž!! = π‘Ž!!  

β€’ π‘Ž!×𝑏! = π‘Žπ‘ !      Solving  by  substitution  Express  the  indices  into  its  component  powers  before  doing  the  substitution    Example:  By  using  a  suitable  substitution,  solve  the  following  equation:    

3!!!! + 3!!! = 1 + 3!  3! 3!! + 3 3! = 1 + 3!  27 3!! + 2 3! βˆ’ 1 = 0  

Let  π‘¦ = 3!  27𝑦! + 2𝑦 βˆ’ 1 = 0  

𝑦 = 0.1589  π‘œπ‘Ÿ βˆ’ 0.2330   𝑁.𝐴  (𝐸π‘₯π‘π‘œπ‘›π‘’π‘›π‘‘π‘–π‘Žπ‘™π‘   π‘Žπ‘Ÿπ‘’  π‘Žπ‘™π‘€π‘Žπ‘¦π‘   π‘π‘œπ‘ π‘–𝑑𝑖𝑣𝑒)  

3! = 0.1589  Take  β€œlg”  both  sides  

π‘₯ lg 3 = lg 0.1589  

π‘₯ =lg 0.1589lg 3

= 0.4771