FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 2.37Example 52 : Simplify a a a a for a 3=^1615
Solution : (Note carefully)
{ } { }
1 1 1 1 1^12 1 1 1/2^12
a a a a a .(a ) (a )=^2 2 2 2 2 (a )2 2^⇒a .a .a .a a^121418116 =^1516when a 3=^1615 ,then a a a a (3 )=^1 6 1515 16= 3
Example 53: If a 2 2=^13 − −^13 show that 2a 6a 3 0^3 + − =
Solution : Given that a 2 2=^13 − −^13
Taking cube of both sides
a 2 2 3(2 2 ) 2^3113131 3a^3 3a
2 2= − − − − − = − − = −
i.e., 2a 6a 3 0^3 + − =
Example 54: if x 5 5 5= −^23 −^18 , prove that
x 15x 60x 20 0^3 −^2 + − =Solution : (^) (5 x) 5 5− =^23 +^13
Cube both sides and simplify,
Example 55: if a bb= a show that
ab ( ) 1a
a ab
b
= −
and if a=2b show that b=2Solution : (^) a b b ab= a⇒ = ba
ab ab ab ( ) 1a
a ab a a ab
b a a
(^) = (^) = = −
(^)
If a=2b then
ab ( ) 1a 2bb ( ) 12b
a ab 2b (2b) b
b b
= − ⇒ = −
⇒4 2b b 2= ⇒ =Example 56: if (1.234) (0.1234) 10a= b= c, show that a c b1 1 1− =
Solution:
(1.234) 10 (0.1234 10) 10a= c a c