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Indices, Surds and Logarithms
Logarithms Common Laws
log! π + log! π = log!(ππ)
log! π β log! π = log!ππ
log! π! = π log! π log! π =
!"#! !!"#! !
(b can take any value) Converting from log from to index Form
log! π = π β π = π! Example:
log! 8 = 3 β 8 = 2! Common & Natural logarithms
log!" π = lg π log! π = ln π
(Use βlnβ when there is π ; otherwise always use βlgβ) Solving Equations 2 Scenarios to reach: 1. A single log on both sides with the same base Example:
log!(2π₯ + 1) = log!(π₯ β 3) 2π₯ + 1 = π₯ β 3
π₯ =β3 β 12
= β2 2. A single log on one side of the equation Example:
log! 2π₯ + 1 = 2 2π₯ + 1 = 3!
π₯ =9 β 12
= 4
Surds Common Laws
π Γ π = ππ ππ=
ππ
π Γ π = π π π Γπ π = ππ ππ
π + π π + π = ππ + π π + π π + ππ 1 + π 1 β π = 1! β π
!= 1 β π
Rationalising the denominator To remove the surd from the denominator, multiply the entire fraction by the conjugate surd Example:
34 β 2 6
=3
4 β 2 6Γ4 + 2 64 + 2 6
=12 + 6 6
4! β 2 6!
=6(2 + 6)16 β 4(6)
= β3(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!Γπ! = π!!! β’ π! Γ· π! = π!!! β’ π!! = !
!!
β’ !!
!= !!
!!
β’ !!
!!= !
!
!= !!
!!
β’ !!!!
!!!!= !!!
!!!
β’ π! = 1 β’ π
!! = π
β’ π!! = π!
β’ π!! = π!!
β’ π!Γπ! = ππ ! 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! β 1 = 0
Let π¦ = 3! 27π¦! + 2π¦ β 1 = 0
π¦ = 0.1589 ππ β 0.2330 π.π΄ (πΈπ₯ππππππ‘ππππ πππ πππ€ππ¦π πππ ππ‘ππ£π)
3! = 0.1589 Take βlgβ both sides
π₯ lg 3 = lg 0.1589
π₯ =lg 0.1589lg 3
= 0.4771
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