MEMO TUTORIAL 1
(1) For each of the following, find an equation of the line satisfying the given conditions.
(a) Passing through the origin and (−2, 3).
The equation is 2y + 3x = 0.
(b) Passing through (−3, 2) and parallel to 2x − y − 3 = 0.
The equation is y = 2x + 8.
(c) Passing through (1, −1) and perpendicular to the y-axis.
The equation is y = −1.
(d) Passing through (1, 4) and perpendicular to x + 3y = 0.
The equation is y = 3x + 1.
(2) Express in term of log 2, log 3 and log 5 the following:
(a) log 10−4 = −4 log 10 = −4 log(5 × 2) = −4 log 5 − 4 log 2.
(b) log 81 = − log 8 = − log 23 = −3 log 2.
(c)
1
8
log 256 =
1
8
log 28 =
8
8
log 2 = log 2.
(d) log 1.25 = log(125 × 10−2 ) = log(125) − 2 log 10 = log(53 ) − 2 log(5 × 2) = log 5 − 2 log 2.
(3) Use the properties of logarithms to express the fol lowing logarithms in terms of logarithms
of x, y and z.
(a) logb (xy 2 ) = logb x + 2 logb y.
(b) logb
√
x2 y
z5
= 2 logb x + 21 logb y − 5 logb z.
(4) Solve the given equation for x.
(a) log8 x + log8 (x + 6) = log8 (5x + 12)
log8 x + log8 (x + 6) = log8 (5x + 12)
log8 (x(x + 6)) = log8 (5x + 12)
x(x + 6) = 5x + 12
x2 + 6x = 5x + 12
x2 + x − 12 = 0
(x − 3)(x + 4) = 0
As possible solution we have 3 and −4. However, the solution is 3 since log8 does not
take negative values.
1
(b) log2 (x + 5) − log2 (2x − 1) = 5
log2 (x + 5) − log2 (2x − 1) = 5
x+5
=5
log2
2x − 1
x+5
= 25
2x − 1
x+5
= 32
2x − 1
(x + 5) = 32(2x − 1)
x + 5 = 64x − 32
37
x= .
63
(c) logx (x + 2) = 2
logx (x + 2) = 2
x + 2 = x2
x2 − x − 2 = 0
(x + 1)(x − 2) = 0
x = 2.
(d) 3x+2 = 7
3x+2 = 7
(x + 2) log 3 = log 7
log 7
− 2.
x=
log 3
(e) log x = 13 (log 16 + 2 log 2)
1
log x = (log 16 + 2 log 2)
3
1
2
= log(2 × 23 ) + log 2)
3
3
1
1
2
= log(2) + log(23 ) + log(2)
3
3
3
3
3
= log 2 + log 2
3
3
= log 2 + log 2
= log 4
x = 4.
(5) Solve the following in R by factorising the expressions.
(a) x5 − 5x3 = 36x = x(x − 3)(x + 3)(x2 + 4), the solution are −3, 0, 3.
√
√
√
√
(b) x2 + 2 2x + 2 = (x + 2)(x + 2), the solution is − 2.
(c) 1 +
2
x
−
8
x2
=
x2 +2x−8
x2
=
(x−2)(x+4)
,
x2
the solution are 2, −4.
2
(d) 4x − 4x2 = 4x(−x + 1), the solution are 0, 1.
(e)
2
x+5
−
x
x−5
=1⇒0=
2
x+5
−
x
x−5
(6) Solve
( the following equations.
− 57 − 11
7 x = −y
(a)
, (−3, −4),
2y = 7 + 5x
(7)
∗
−1=
−2x2 −3x+15
(x−5)(x+5) .
Therefore, x =
(
2y 2 − x2 = 1
(b)
x − 2y = 3.
√
3± 129
−4
, (1, −1), (−7, −5).
Let n be a nonzero integer. Compute the sum
1
1
S = log(an ) + log(an−1 ) + · · · + log(a) + log(1) + log( ) + · · · + log( n ).
a
a
an · · · a
= log( n
)
a ···a
= log 1
= 0.
(8)
∗∗
Consider a rectangle with perimeter 28 cm and diagonal 10 cm. Find the length and
width of the rectangle. Let x be the length
p and y the width. Then we have x > y, and the
perimeter is 2x + 2y and the diagonal is x2 + y 2 .
(
2x + 2y = 28
x2 + y 2 = 102 .
We get (x, y) = (8, 6)
3