EXPONENT &LOGARITHMS
MORE EXAM STYLE QUESTIONS
2x
1. Solve the equation 9x–1 =  1  .
 3
2.
If loga 2 = x and loga 5 = y, find in terms of x and y, expressions for
(a)
log2 5;
(b)
loga 20.
2
3.
 P 
 in terms of x , y and z.
Let log10P = x , log10Q = y and log10R = z. Express log10 
3 
 QR 
4. Solve the equation log27 x = 1 – log27 (x – 0.4).
5. Given that log5 x = y, express each of the following in terms of y.
(a)
log5 x2
(b)
1
log5  
 x
(c)
log25 x
6.It is thought that a joke would spread in a school according to an exponential model
N = 4 × (1.356)0.4t, t  0; where N is the number of people who have heard the joke, and t is
the time in minutes after the joke is first told.
(a)
How many people heard the joke initially?
(b)
How many people had heard the joke after 16 minutes?
There are 1200 people in the school.
(c)
Estimate how long it would take for everybody in the school to hear this joke.
ANSWERS
2x
x–1
1. 9
1
= 
 3
2x–2
3
= 3–2x
2x – 2 = –2x
1
x=
2
(M1) (A1)
(A1)
(A1) (C4)
[4]
2.
(a)
log2 5 =
=
(b)
log a 5
log a 2
(M1)
y
x
(A1) (C2)
loga 20 = loga 4 + loga 5 or loga 2 + loga 10
= 2 loga 2 + loga 5
= 2x + y
(M1)
(A1) (C2)
[4]
2
 P 


 = 2log10  P 
3.
log10 
3 
3 

 QR 
 QR 
 P 
 = 2(log10P – log10(QR3))
2log10 
3 
 QR 
= 2(1og10P – log10Q – 3log10R)
= 2(x – y – 3z)
= 2x – 2y – 6z or 2(x – y – 3z)
(M1)
(M1)
(M1)
(A1)
[4]
4.
log27 (x(x – 0.4)) = l
x2 – 0.4x = 27
x = 5.4 or x = –5
x = 5.4
(M1)(A1)
(M1)
(G2)
(A1) (C6)
Note: Award (C5) for giving both roots.
[6]
5.
(a)
log5 x2 = 2 log5 x
= 2y
(b)
log5
(c)
1
= –log5 x
x
= –y
log25 x =
=
(M1)
(A1) (C2)
(M1)
(A1) (C2)
log 5 x
log 5 25
(M1)
1
y
2
(A1) (C2)
[6]
6.
(a)
4
(A1) (C1)
(b)
For raising to a power of 6.4
28
(M1)
(A1) (C2)
(c)
1200 = 4 × (1.356)0.4t (for substituting in the formula)
300 = (1.356)0.4t
t = 46.8 (by trial and error)
(M1)
(A1)
(A1)
OR
t = 46.8
(G3) (C3)
[6]