Quantitative aptitude

2. 1/2 log a to the base b

     3log b to the base c

      3/2 log ato the base c

1. log a / log b, log b / log c, log c / log b

4. log t + log (t-3)=1

or, log(2t-3-1) = 0

or, log(2t-4) = 0

or, 2t-4=0

or,2t = 4

or, t = 2.ans. 

4.) log(t*(t-3)) = 1 implies t^2 -3*t = 10 implies t^2 - 3t - 10 = 0. solving the quadratic we get t = 5 or -2 logarithm of negative number is complex entity so we discard t = -2 t =5 is the answer verification: log(5) + log(5-3) = log(5) + log(2) = log(10) =1

3.) all the logarithms have the same base.

log(b*2b^2*3b^3*.... *nb^n) = log(n! * b^(1+2+3+....+n)) = log(n!) + (n(n+1)/2)*log(b) (all the logarithms are to base a)

2.) loga to base b can be wrritten as (loga to base 10)/(logb to base 10) now getting the power outside the answer is 1/2(loga)/(logb) * 3*(logb)/(logc) * 3/2(logc)/(loga) = 9/4

1. similar to second problem In this the second and third terms cancel each other so you are left with loga to base b