Q. No. 7: If loga (ab) = x, then logb (ab) is
A : 1/x
B : x/(x+1)
C : x/(x-1)
D : x/(1-x)
Answer: C loga (ab) = x => 1 + logb/loga = x Thus, logb/loga = x-1and loga/logb = 1/(x-1) logb (ab) = 1 + loga/logb = 1 + 1/(x-1) = x/(x-1)
Q. No. 8: If logx a, ax/2 and logb x are in GP, then x is
A : loga (logb a)
B : loga (loge a) + loga (loge b)
C : -loga (loga b)
D : loga (loge b) - loga (loge a)
Answer: A logx a , ax/2 and logb x are in GP Then, (ax/2 )2 = logx a * logb x => ax = logb a =>x log a = log (logb a) Hence, x = loga (logb a)
Q. No. 9: 1/2*(log10 25) - 2(log10 3) + log10 18 equals
A : 18
B : 1
C : log10 3
D : 2
Answer: B =>
log10 (25)1/2 - log10 (3)2 + log10 18 =>log10 5 - log10 9 + log10 18 = log10 (5/9) + log10 18 => log10 (5*18/9) = log10 10 =1
Q. No. 10: The difference between the logarithms of sum of the squares of two positive numbers A and B and the sum of logarithms of the individual numbers is a constant C. If A = B, then C is
A : 2
B : 1.3031
C : log 2
D : exp(2)
Answer: C Given, log (A2 + B2 ) - (log A + log B) = C If A = B, then log (2 A2 ) - 2 log A = C log(2 A2 ) - log(A2 ) = C log (2 A2 /A2 ) = C => C = log 2
Q. No. 11: Which of the following is True?
A : log17 275 = log19 375
B : log17 275 < log19 375
C : log17 275 > log19 375
D : None of the above
Answer: B log17 275 = log19 375 => log 275/log 17 = log 375/log 19 => 275/17 = 375/19, therefore 16.18 < 19.74 Hence, log17 275 < log19 375
Q. No. 12: If log12 27= a, then log6 16 is
A : 1/4 *(3-a)/(3+a)
B : 1/4 * (3+a)/(3-a)
C : 4* (3+a)/(3-a)
D : 4*(3-a)/(3+a)
Answer: D log12 27 = a => log12 3 = a/3 log3 12 = 3/a => log3 4 + log4 3 =3/a => 2log 32 = (3-a)/a => log3 2= (3-a) /2a......(i) Now, log6 16= 4log6 2= A => log3 2= A/(A-4)..........(ii) From (i) and (ii), A= 4(3-a)/(3+a)