Questions on Logarithmic Functions and Indices

1.            Solve the following equations:

a)      i) 3^2x = 1000             (Cambridge IGCSE A-Maths 0606/23, May/June 2012 P2 Q5)

      ii) 36^2y-5/6^3y = 6^2y-1/216^y+6 

b)      4^x / 2^5-x = 2^4x / 8^x-3   (IGCSE Q)

c)      3^x+1 + 3^2-x = 28    (IGCSE Q)

d)      2^3x = 8 + 2^{3x – 1}   (SPM 2011 P1 Q7 @ pg 162)

e)      3^{x + 2} – 3^x = 8/9    (SPM 2010 P1 Q7 @ pg 136)

f)        16^2x-3 = 8^4x            (SPM 2008 P1 Q7 @ pg 83)

g)      9(3^{n – 1}) = 27ⁿ      (SPM 2007 P1 Q8 @ pg 58)

h)      3^{n – 3} x 27ⁿ =243   (SPM 2009 P1 Q7 @ pg 110)

i)        2^{x + 4} – 2^{x + 3} = 1   (SPM 2005 P1 Q7 @ pg 5)

j)        8^2x-3 = 1/√(4^{x + 2})  (SPM 2006 P1 Q6 @ pg 31)

k)      27(3^{2x + 4}) = 1     (SPM 2012 P1 Q7 @ pg 187)

2.      (i)  Write 125 = 5³ as a logarithmic equation

(ii) Express log_b32 = 5 in index form

 (iii) Evaluate:

(a)      b^log_b^x 

(b)     10^lg10

(c)      e^{ln x} for x = 3                                   

3.      For each of the following equations, sketch on separate axes its graph, showing clearly where the graph crosses the axis:

a)      y = e^x                                                                           (2)

b)      y = log₃ x                                                                      (2)

c)      y = 2^-x                                                                          (2)

d)      y = log₃ (-x)                                                                  (2)

                                      

4.      Solve the equations:

a)      log_x 128 = 7                                                                 (2)

b)      log₅ (7x -1) = 3                                                            (3)

c)      log₄ t = 6 log_t 16 -1                                                      (5)

5.      Solve

(a)   log_q 343 = 3                                                                             (2)

(b)   log₄ (5n + 9) = 3                                                                       (3)

(c)   log_m 4 + 8 log₄ m = 6                                                                (6)

(d)   2 log₃ x – 3x log₃ x + 6x = 4                                                     (5)

6.      Solve

(a) log_x 125 = 3                                                                             (2)

(b) log₄ (9y + 4) = 4                                                                       (3)

(c) 3 – log_p p = log_p 9                                                                     (6)

7.      Given that f(x) = log₅ 3 + log₅ 6 + log₅ 9 + log₅ 12 + log₅ 15

a)      Show that f(x) = 6 log₅ 3 + 3 log₅ 2 + 1                        (3)

b)      Solve f(x) = 1 + log₅ x + log₅ x²                                    (3)

8.      a) Given that log₈ 7 = m log₂ 7, find the value of m.                          (2)

      b) f(x) = 16x log₉ 10 – 12 log₉ 10 + 4x log₃ x – 3 log₃ x

i)        Factorise f(x) completely                                                     (3)

ii)   Hence, solve f(x) = 0                                                          (3)

9.       Solve the simultaneous equations:                                                  

(a)    2 log₃ x + 3 log₅ y = 7

                                 log₃ x – log₅ y = 1                                                (4)

(b)   (Given that p ≠ q):  log_p q + 3 log_q p = 4

                                          pq = 81                        (5)

(c)    3 log₂ x + 4 log₃ y = 10

   log₂ x – log₃ y = 1                                                (6)

10.  Solve:         (a) log_q 5 + 6 log₅ q = 5                                                (4)

                  (b) log₃ (5x + 12) + log₃ x = 2                                      (5)

11.  Given that log₉ 10 = k log₃ 10,

(a)   find the value of k.                                                                     (2)

      (b) Factorise completely

                                    4x log₃ x – 3 log₃ x + 16x log₉ 10 – 12 log₉ 10  (2)

(c)   Hence, solve the equation                                                          (5)

                        4x log₃ x – 3 log₃ x + 16x log₉ 10 – 12 log₉ 10 = 0       

                       

12.    Solve

(a)    log_p 343 = 3                                                                (2)

(b)   log₆(11q – 4) = 3                                                                     (2)

(c)    9 log_r 3 = log₃ r                                                                        (3)

(d)   Show that                                                                                (3)

           log₅ 3 + log₅ 6 + log₅ 9 + log₅ 12 + log₅ +15 = 6 log₅ 3 + 3 log₅ 2 + 1

(e)    Solve the equation                                                                   (3)

           log₅ 3 + log₅ 6 + log₅ 9 + log₅ 12 + log₅ +15 = 1 + log₅ x + log₅ x²

13.    (a) Solve the equations

(i)                  log_x 343 = 3                                                              (2)

(ii)                log₉(4y – 3) = 2                                                         (2)

(b)   Solve, to 3 significant figures, log_q 5 + 6 log₅ q = 5        (5)

(c)    Show that x log₂ x⁵ – log₂ x² ≡ (5x – 2) log₂ x.               (2)

(d)   Hence solve the equation x log₂ x⁵ – log₂ x² = 20x – 8   (4)

                                                           

14.    (a) Solve the equations log₄ 2 = p                                       (1)

Given that log₂ 3 = k log₄ 3

(b)   find the value of k                                                                   (2)

(d)      Show that                                                                               (4)

     5x log₄ x – 2 log₄ x – 10x log₂ 3 + 4 log₂ 3 = log₄ (x^{5x – 2}/3^{20x – 8})

(e)      Hence solve the equation                                                        (4)

5x log₄ x – 2 log₄ x – 10x log₂ 3 + 4 log₂ 3 = 0

15.    Solve

(a)    log_q 343 = 3                                                    (2)

(b)   log₄(5n + 9) = 3                                               (3)

(c)    log_m 4 + 8 log₄ m = 6                                       (6)

(d)   2 log₃ x - 3x log₃ x + 6x = 4                             (5)

16.  Solve

(a)      log_p 243 = 5                                                               (2)

(b)      log₄ (3q + 4) = 3                                                        (2)

f(x) = 2x log_x 3 – 5 log_x 9 – x + 5

(c)      Find the value of a and the value of b so that

         

          f(x) = (x – 5)(a log_x 3 – b)                                (3)

(d)      Hence solve the equation f(x) = 0                                (3)

17.    (a) Solve the equations

(i)                  log₅ 625 = x                                                              (2)

(ii)                log₃(5y + 3) = 5                                                         (2)

(b)   (i)       Factorise          5x ln x + 3 ln x – 10x – 6                     

(ii)      Hence find the exact solution of the equation

5x ln x + 3 ln x – 10x – 6 = 0                            (5)

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Simultaneous Equations - Edexcel Past-Year Questions

Simultaneous Equations – Past-Year Questions:

1.        Solve the simultaneous equations

3x + y = 1

   x² + 3xy + y² = 5                                     (6)                    …24/03/2004, P1, Q4

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2.        Solve the simultaneous equations

                    2x² + 9xy + 9y² = 0

                    3x + 3y = 2                                          (6)                    …27/05/2004, P1, Q2

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3.        Solve the simultaneous equations

             y = x² – 7x + 10

             x – y + 3 = 0                                              (5)                    …26/01/2005, P1, Q3

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4.        Solve the equations

y – 2x = 5

     2x² + 2xy + y² = 5                                (6)                    …25/05/2005, P1, Q2

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5.        Solve the equations

                        x² + 4x – xy = 10

                 2x – y = 3                                            (6)            …24/05/2006, P1, Q1

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6.        Find the coordinates of the points of intersection of the curve with equation y = 3x² – 4x + 2 and the line with equation 7x + y = 8.        (5)                 …19/01/2007, P1, Q3

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7.        (a)  Find the coordinates of the points where the line with equation y = 3x + 3 intersects the

       curve with equation y = x² + x – 12.                                                                  (5)

(b)   Find the set of values of x for which x² + x – 12 ≥ 3x + 3.                                 (2)

                                                                                                  …17/5/2007, P2, Q4

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8.        Find the coordinates of the points of intersection of the curve with equation y = x² – 8x + 11 and the line with the equation x + y = 5.         (5)         …12/5/2008, P1, Q2

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9.         Solve the equations

x – 2y = 3

     2y² + 2xy + x² = 1                                (6)                       …13/5/2010, P1, Q3

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10.    Solve the equations

    xy = 6

     xy + x + y = 11                                    (6)                          …15/12010, P1, Q4

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11.    Find the coordinates of the points where the line with equation y = 2x - 5 meets the

curve with the equation xy = 12.                 (5)                                …14/5/2009, P2, Q11            

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12.    The grid opposite shows the graph of y = 3x – 4 + 6/x²

The line with equation y = 5x – 4 intersects the curve with equation y = 3x – 4 + 6/x²

(a)    Using algebra, show that the x-coordinate of P satisfies x³ = 3.          (3)

(b)   By drawing a suitable straight line on the grid, obtain an estimate, to 1 decimal place, for the value of 3^1/3                                                                                                                                       
                                                                                                           (13/05/2010, P1, Q1)
                                                                                (+ many more such questions)

                                                                                             

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Trigonometric Equations Involving Addition Formula

Solving Trigonometric Equations Involving Addition Formula

 Q1) Using cos(A + B) ≡ cos A cos B – sin A sin B,

(a)    show that

(i)    sin² Ө = (1/2)(1 – cos 2Ө),

(ii)   cos² Ө = (1/2)(cos 2Ө + 1)                                                            (4)

f(Ө) = 1 + 10 sin² Ө - 16 sin⁴ Ө .

(b)   Show that f(Ө) = 3 cos 2Ө – 2 cos 4Ө.                                               (4)

(c)    Solve the equation

1 + 10 sin² Ө^o - 16 sin⁴ Ө^o + 2 cos 4Ө^o = 0.25, for 0 ≤ Ө ≤ 180

             giving your solutions to 1 decimal place.                                       (4)  

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Possible Solutions:

Given: cos (A + B) ≡ cos A cos B – sin A sin B            …Eqn (1)

a)      To show:

(i)                  sin² Ө = (1/2)(1 – cos 2Ө)                    …Eqn (2)

RHS = (1/2)(1 – cos 2Ө) = (1/2)[1 – cos (Ө + Ө)]

        = (1/2)[1 – (cos² Ө - sin² Ө)] = (1/2)[1 – (cos² Ө) + sin² Ө]

Use identity: cos² Ө = 1 - sin² Ө

RHS = (1/2)[1 – (1 - sin² Ө) + sin² Ө]

        = (1/2)[1 – 1 + sin² Ө + sin² Ө]

        = (1/2)[2 sin² Ө]

        = sin² Ө = LHS

(ii)                cos² Ө = (1/2)(cos 2Ө + 1)                  …Eqn (3)

RHS = (1/2)(cos 2Ө + 1) = (1/2)[cos (Ө + Ө) + 1]

        = (1/2)[(cos² Ө - sin² Ө) + 1] = (1/2)[1 + cos² Ө – (sin² Ө)]

Use identity: sin² Ө = 1 - cos² Ө

RHS = (1/2)[1 + cos² Ө – (1 - cos² Ө)]

        = (1/2)[1 + cos² Ө – 1 + cos² Ө]

        = (1/2)[2 cos² Ө]

        = cos² Ө = LHS

Given: f(Ө) = 1 + 10 sin² Ө – 16 sin⁴ Ө                        …Eqn (4)

b)      To show: f(Ө) = 3 cos 2Ө – 2 cos 4Ө              …Eqn (5)

Fr. (4):             f(Ө) = 1 + 10 sin² Ө – 16 sin⁴ Ө

                        f(Ө) = 1 + 2 sin² Ө (5 – 8 sin² Ө)

Use (2):            f(Ө) = 1 + 2 [(1/2)(1 – cos 2Ө)][5 – 8 ((1/2)(1 – cos 2Ө))

                        f(Ө) = 1 + (1 – cos 2Ө)[5 – 4(1 – cos 2Ө)]

                        f(Ө) = 1 + (1 – cos 2Ө)[5 – 4 + 4 cos 2Ө)]

                        f(Ө) = 1 + (1 – cos 2Ө)(1 + 4 cos 2Ө)

                        f(Ө) = 1 + (1 + 3 cos 2Ө - 4 cos² 2Ө)

Use (3):            f(Ө) = 1 + [1 + 3 cos 2Ө - 4 ((1/2)(cos 4Ө + 1))]

                        f(Ө) = 1 + [1 + 3 cos 2Ө - 2(cos 4Ө + 1)]

                        f(Ө) = 1 + [1 + 3 cos 2Ө - 2cos 4Ө - 2]

                        f(Ө) = 1 + 1 + 3 cos 2Ө - 2cos 4Ө - 2

                        f(Ө) = 3 cos 2Ө – 2 cos 4Ө  (Shown Eqn 5 from Eqn 4)

c)      To solve, for 0 ≤ Ө ≤ 180:

1 + 10 sin² Ө^o - 16 sin⁴ Ө^o + 2 cos 4Ө^o = 0.25            …Eqn (6)

Fr. (5):              2 cos 4Ө = 3 cos 2Ө – f(Ө)                             …Eqn (7)

Put (7) and (4) into (6): (by “double substitution”)

                                   

                                    f(Ө) + (3 cos 2Ө – f(Ө)) = 0.25

                                                            3 cos 2Ө = 0.25

                                                   cos 2Ө = (0.25/3)

Domain for 2Ө: 2 x 0 ≤ 2Ө ≤ 2 x 180 or  0 ≤ 2Ө ≤ 360

Since cos 2Ө is +ve:     2Ө is in 1^st Qd or 4^th Qd

                                    2Ө = cos^-1 (0.25/3)      …1^st Qd

                                     Ө = (1/2) cos^-1 (0.25/3)

                                     Ө = 42.6099…

                                     Ө = 42.6 (1 dp)

                        Or,       2Ө = 360 - cos^-1 (0.25/3)         …4th Qd

                                     Ө = (1/2) [360 - cos^-1 (0.25/3)]

                                     Ө = (1/2) (274.78019..)

                                     Ө = 137.390…

                                     Ө = 137.4 (1 dp)

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Solving Simultaneous Equations - Double Substitution Method

The following simultaneous equations may be quickly and easily solved by “double-substitution” method.

The conventional single substitution method is often times too lengthy and very prone to errors. Try your conventional method, before looking at  my "double substitution' method and you will know what I mean.

So, always have a quick and closer look at the simultaneous equations before you - if they are amenable to “double-substitution” method, then use it. (Now no longer in Form 2 or Form 3, so use a more "skillful method" where possible. hahaha  :))

1.        Solve the simultaneous equations

3x + y = 1

x² + 3xy + y² = 5 

                                                                                                        

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2.        Solve the simultaneous equations

              2x² + 9xy + 9y² = 0

              3x + 3y = 2

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The “Double-Substitution” Method:

Q 1)                             3x + y = 1                    …Eqn (1) 

x² + 3xy + y² = 5          …Eqn (2)

Fr. (1):                y = 1 – 3x                    …Eqn (3)

Fr. (2):                x² + y(3x + y) = 5         …Eqn (4)

Put (1) & (3) into (4):

                           x² + (1 – 3x)(1) = 5

                           x² + 1 – 3x = 5

                           x² – 3x – 4 = 0

                           (x + 1)(x – 4) = 0

                           x = -1                           or,        x = 4

Use (3):              y = 1 – 3(-1)                or,        y = 1 - 3(4)

                           y = 4                            or,        y = -11

Answer: x = -1, y = 4

               x = 4, y = -11

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Q2)                     2x² + 9xy + 9y² = 0      …Eqn (1)

3x + 3y = 2             …Eqn (2)

Fr. (2):                3y = 2 – 3x                  …Eqn (3)

Fr. (1):                2x² + 3y(3x + 3y) = 0 …Eqn (4)

Put (2) & (3) into (4):

                           2x² + (2 -3x)(2) = 0

                           2x² + 4 – 6x = 0

                           2(x² + 2 – 3x) = 0

                           2(x - 1)(x – 2) = 0

                           x = 1                            or,        x = 2

Use (3):              3y = 2 – 3(1)                or,        3y = 2 – 3(2)

                           y = -1/3                        or,        y = -4/3

Answer: x = 1, y = -1/3

               x = 2, y = -4/3

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You can also use 'double substitution' to solve part (c) of the question below easily. Try your conventional method first before you click here for my "double-substitution" method.

1)        Using cos(A + B) ≡ cos A cos B – sin A sin B,

(a)    show that

(i)             sin² Ө = (1/2)(1 – cos 2Ө),

(ii)           cos² Ө = (1/2)(cos 2Ө + 1)                                                            (4)

f(Ө) = 1 + 10 sin² Ө - 16 sin⁴ Ө .

(b)   Show that f(Ө) = 3 cos 2Ө – 2 cos 4Ө.                                               (4)

(c)    Solve the equation

1 + 10 sin² Ө^o - 16 sin⁴ Ө^o + 2 cos 4Ө^o = 0.25, for 0 ≤ Ө ≤ 180

             giving your solutions to 1 decimal place.                                               (4) 

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