## anonymous one year ago log abc^0.5+log abc =?

1. anonymous

options are: 5,9,2,1

2. xapproachesinfinity

let me see if this what you wrote before we begin $\log abc^{\frac{1}{2}}+\log abc$

3. xapproachesinfinity

is that right?

4. xapproachesinfinity

is it (abc)^0.5 or just ab(c^0.5) huge difference

5. xapproachesinfinity

so which is it?

6. xapproachesinfinity

hello!! you there?

7. anonymous

yes the first one is ryt

8. anonymous

will be 1/2log abc +log abc

9. xapproachesinfinity

so it is (abc)^1/2 yes

10. anonymous

yes

11. xapproachesinfinity

okay then $$\log (abc)^{frac{1}{2}}+\log abc= \frac{1}{2} \log abc+\log abc$$ $$=\frac{3}{2}\log(abc)$$

12. xapproachesinfinity

that first line where frac12 supposed to be all to 1/2

13. xapproachesinfinity

so we get $$=\log(abc)^{\frac{3}{2}}$$

14. anonymous

yea , next step

15. xapproachesinfinity

that is all what is it you want to achieve?

16. xapproachesinfinity

a different way of doing that is $$\log (abc)^{\frac{1}{2}}+\log abc=\log \frac{(abc)^{\frac{1}{2}}}{abc}$$

17. anonymous

but the options of this ques given are a) 5 b) 9 c) 2 d) 1 which onw wud be the ans?

18. xapproachesinfinity

i meant $$=log[(abc)^0.5 (abc)]$$ not division!!

19. xapproachesinfinity

$$=\log[(abc)^{0.5} (abc)]$$

20. xapproachesinfinity

if those are your options then you missing something in the question

21. anonymous

alright

22. xapproachesinfinity

can you post the picture of your question

23. anonymous

that must be the wrong ques. i pick this ques from mformaths.

24. anonymous

bdw thankew for your help. :)

25. xapproachesinfinity

that question cannot be any of those number no more simplification is required

26. anonymous

yep

27. xapproachesinfinity

all you can do is something like this $$=3/2[\log a+\log b+ \log c]$$ which pretty much does not help for anything new

28. anonymous

ok i have another question cud u help me in that?

29. xapproachesinfinity

post is in a different post though

30. anonymous

okz