Question

Another log-
x=log(basea)bc , y=log(baseb)ca , z=log(basec)ab
then 1+abc = ?
1) x+y+z 2)(1+x)^{-1} + (1+y)^{-1} + (1+z)^{-1}
3) xyz 4)None of the above.

Answer 1

IF\[x=\log_{a}bc \]\[y=\log_{b}ca \]\[z=\log_{c}ab \]
then 1+abc=?

Answer 2

like one opinion

1. - so i think that you need to know the property of logharitms
2. - you need to rewriting this x,y and z in the same base of logharitm
3. - rewrite ,,1 +abc = " using these terms what you have got after the bases was changed in the same of every logarithms
4. - check the resulte and see how can using there these terms of x,y and z like final expression

- hope so much that is understandably and that i have helped you anything with these ideas

good luck

bye

Answer 3

@shubhamsrg check options.

Answer 4

2) is the answer.How??

Answer 5

@shubhamsrg are you solving??

Answer 6

shubham i think u r ryt

Answer 7

yes i am,,gimme 1 min,,will just pots it

Answer 8

post*

Answer 9

ok

Answer 10

guys i confused between u u got the same name

Answer 11

sorry am unable to do as of now
gimme more time,,hmm

Answer 12

ok

Answer 13

you sure ans is 2 ?

Answer 14

cause am getting (1+x)^{-1} + (1+y)^{-1} + (1+z)^{-1} =1
which indirectly means either of a,b,c is 0,for which log wont be defined