Worked example 1: Applying the exponential laws
QUESTION
Simplify:
- 23 x 24 x
2.
4 x^3
2 x^53.12 p^2 t^5
3 pt^3
4.(3x)^2
5.(
3452
) 3
- 6 p^0 (7p)^0
7.(
2 xp
6 x^2) 3
8.
(
2 ^2
) 2 x+1SOLUTION
- 23 x 24 x= 2^3 x+4x= 2^7 x
2.4 x^3
2 x^5= 2x^3 ^5 = 2x ^2 =2
x^2
3.
12 p^2 t^5
3 pt^3= 4p(2 1)t(5 3)= 4pt^24.(3x)^2 = 3^2 x^2 = 9x^2
5.(
34 52
) 3
= 3(43) 5 (23)= 3^12 56
- 6 p^0 (7p)^0 = 6(1)1 = 6
7.(
2 xp
6 x^2) 3
=
(p
3 x) 3
=
p^3
27 x^3
8.(
2 ^2
) 2 x+1
= 2 2(2x+1)= 2 ^4 x ^2NOTE:
When you have a fraction that is one term over one term, use the method of Finding Prime Bases - in other
words use prime factorisation on the bases.Worked example 2: Exponential expressions
QUESTION
Simplify:
22 n 4 n 2
16 n
SOLUTION
Step 1: Change the bases to prime numbers
At first glance it appears that we cannot simplify this expression. However, if we reduce the bases to prime
bases, then we can apply the exponent laws.
22 n 4 n 2
16 n=
22 n(
22
)n
21
(2^4 )n46 2.2. Revision of exponent laws