If log 2, log (2x-1) and log (2x+3) are in AP, show that x= log5/log2.
Answers
Note :
★ A.P. (Arithmetic Progression) : A sequence in which the difference between the consecutive terms are equal is said to be in A.P.
★ If a1 , a2 , a3 , . . . , an are in AP , then
a2 - a1 = a3 - a2 = a4 - a3 = . . .
★ The common difference of an AP is given by ; d = a(n) - a(n-1) .
★ The nth term of an AP is given by ;
a(n) = a1 + (n - 1)d .
★ If a , b , c are in AP , then 2b = a + c .
Solution :
★ Given : log2 , log(2x-1) , log(2x+3) are in AP .
★ To prove : x = 5/2
Since ,
log2 , log(2x-1) , log(2x+3) are in AP
Thus ,
=> 2log(2x-1) = log2 + log(2x+3)
=> log(2x-1)² = log2(2x+3)
=> (2x - 1)² = 2(2x + 3)
=> (2x)² - 2•2x•1 + 1² = 2•2x + 2•3
=> 4x² - 4x + 1 = 4x + 6
=> 4x² - 4x - 4x + 1 - 6 = 0
=> 4x² - 8x - 5 = 0
=> 4x² - 10x + 2x - 5 = 0
=> 2x(2x - 5) + (2x - 5) = 0
=> (2x - 5)(2x + 1) = 0
=> x = 5/2 , -1/2
=> x = 5/2 (appropriate value)
Here ,
x = -1/2 is rejected value .
Because , If x = -1/2 then
log(2x - 1) = log[2•(-½) - 1]
= log(-1-1)
= log(-2) (not defined)