Question

If, a=log(base24)12, b=log(base36)24, c=log(base48)36,
then 1+abc=?
a) 2ab c)2bc
b) 2ac d) 0

Answer 1

@nitz

Answer 2

\[a=\log_{24}12\]\[b=\log_{36}24\]\[c=\log_{48}36\]

Answer 3

this is similar to previous problem

Answer 4

still, help.

Answer 5

abc=log(base 48) 12

Answer 6

i am getting this

Answer 7

me too got the same.

Answer 8

maybe change every base to 12
i.e. a=1/log24 ,,b = log24/log36 ,c= log36/log48
=>abc = 1/log48
1+ abc = 1 + 1/log48 = 1/log12 + 1/log48 = log(48*12)/log48 = log(12^2) + log(4) / log48 = 2 + 2log2 /log48 = 2(1 + log 2)/log48 = 2( log 24)/log48 = 2bc
phew!!
:P

Answer 9

cant say but still i am trying

Answer 10

did you get shubhamsrg method???????

Answer 11

wait.

Answer 12

@shubhamsrg "1 + 1/log48 = 1/log12 + 1/log48 " . How?

Answer 13

as log12 = 1
recall that base is 12 everywhere..

Answer 14

ok thanks.
I got it.