rajiv (student) (197 Points)
23 May 2012Dear friends, please solve the following problems
1. What is the value of Loga to base b . Logb to base c . Logc to base b ?
2. What is the value of Log ba1/2 . Logcb3 . Logac3/2 ?
3. The sum of the series Logab + Loga2b2 + Loga3b3 + ….+ Loganbn is given by ...........
4. On solving the equation Log t + Log (t-3) = 1, we get the value of t as ……………
intermediate(ipc)course
(no)
(1460 Points)
Replied 23 May 2012
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
intermediate(ipc)course
(no)
(1460 Points)
Replied 23 May 2012
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.
Tejaswi Kasturi
(student-cpt)
(427 Points)
Replied 24 May 2012
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
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