log p log a p log a log p a a log a p b -log p a c log

ans is a) log a p
11 years agoHelpfull: Yes(9) No(5)
Ans: B
==> log p (1/log p a)/log a (log p a) let log p a = x
==> log p (1/x)/log a (x)
==> log p (x^-1)/log a (x)
==> - log p (x) /log a (x)
==> - log x (a) /log x (p)
==> - log p (a) >
11 years agoHelpfull: Yes(7) No(0)
ans is log a p
log p (log a p)/log a (log p a)=
(log p(log a/log p))/(log a(log p/log a))=log a p
11 years agoHelpfull: Yes(3) No(4)
mr. nikhil can you clear in (log a p), a is base or a*p.....plz...
11 years agoHelpfull: Yes(1) No(1)
yes this is minus of log a with base p...
option b
11 years agoHelpfull: Yes(1) No(0)
Hi Ashish,

log a p means log p to base a,
11 years agoHelpfull: Yes(0) No(1)
a1 = lg2
a2 = lg(2^x - 1)
a3 = lg(2^x + 3 )
If the given sequence is an a.p., then:
a2 - a1 = a3 - a2
lg(2^x - 1) - lg2 = lg(2^x + 3 ) - lg(2^x - 1)
We'll use the quotient property of the logarithms:
lg [(2^x - 1)/2] = lg [(2^x + 3 )/(2^x - 1)]
Because the bases of logarithms are matching, we'll use the one to one property:
[(2^x - 1)/2] = [(2^x + 3 )/(2^x - 1)]
We'll remove the brackets and cross multiply:
(2^x - 1)^2 = 2*(2^x + 3 )
We'll expand the square from the left side:
2^2x - 2*2^x + 1 = 2*2^x + 6
We'll move all terms to one side:
2^2x - 4*2^x - 5 = 0
We'll substitute 2^x = t
t^2 - 4t - 5 =0
We'll apply the quadratic formula:
t1 = [4+sqrt(16+20)]/2
t1 = (4+6)/2
t1 = 5
t2 = (4-6)/2
t2 = -1
Since 2^x>0, the only admissible solution is:
2^x = 5
x = log 2 (5)
11 years agoHelpfull: Yes(0) No(3)

log p (log a p)/log a (log p a)=
a)log a p b)-log p a c)log a p d)1