SOLUTION
Step 1: Rewrite the equation
In order to get the equation into a form which we can factorise, we need to rewrite the equation:
2 x 24 x= 0
2 x 24 : 2 x= 0
2 x
24
2 x
= 0
Now eliminate the fraction by multiplying both sides of the equation by the denominator, 2 x.
(
2 x
24
2 x
)
2 x= 0 2 x
22 x 16 = 0
Step 2: Factorise the equation
Now that we have rearranged the equation, we can see that we are left with a difference of two squares.
Therefore:
22 x 16 = 0
(2x 4)(2x+ 4) = 0
2 x= 4 2x̸= 4 (a positive integer with an exponent is always positive)
2 x= 2^2 = 4
x= 2
Thereforex= 2.
Exercise 2 – 3:
1.Solve for the variable:
a) 2 x+5= 32 b) 52 x+2=
1
125
c) 64 y+1= 16^2 y+5 d) 39 x ^2 = 27
e) 25 = 5 z ^4 f) ^12 : 6
m 2 +3
= 18
g) 81 k+2= 27k+4 h) 251 ^2 x 54 = 0
i) 27 x 9 x ^2 = 1 j) 2 t+ 2t+2= 40
k) (7x 49)(3x 27) = 0 l)(2: 2 x 16)(3x+1 9) = 0
m) (10x 1)(3x 81) = 0 n) 2 52 x= 5 + 5x
o) 9 m+ 3^3 ^2 m= 28 p)y 2 y
(^12)
- 1 = 0
q) 4 x+3=0,5 r) 2 a=0,125
Chapter 2. Exponents 55