## anonymous one year ago 49log2*2log7 =? please do a favour to solve this ques.

1. rajat97

it is in the form xlogy*ylogx

2. rajat97

what can you do to simplify it?

3. Loser66

.

4. anonymous

|dw:1436709166375:dw|

5. anonymous

can i do somthing like that

6. Loser66

I don't think you can do anything with the original one unless using calculator. It is simplest form since (2,7) are relative prime.

7. anonymous

oke

8. anonymous

49log2*2log7=7^2log2*2log7

9. anonymous

what wud be the next step.

10. SolomonZelman

you are multiplying logarithms (correct?) $$\large\color{black}{ \displaystyle 49\log(2)~\cdot~2\log(7) }$$ At first I will ask you, are you sure that it is supposed to be a "•", and not a "+" ?

11. anonymous

yes it is sure http://www.m4maths.com/placement-puzzles.php?SOURCE=amcat&TOPIC=Numerical%20Ability&SUB_TOPIC=Ratio%20and%20Proportion you can check out at this link

12. SolomonZelman

I don't have a login to this thing, but I of course believe you. I was just asking that for a clarification.....

13. anonymous

oke but i wanna ask is there any formula to solve the above equation , which i hve done by drawing. log 2^7^2

14. SolomonZelman

I think that you can't make it any simpler than $$\large\color{black}{ \displaystyle 49\log(2)~\cdot~2\log(7) =98\log(2)\log(7)}$$ (And of course the assumed base is 10, not e)

15. anonymous

yes!

16. SolomonZelman

Assumed base is also not important tho' for this multiplication, because the assumed base wouldn't matter with this multiplication.

17. SolomonZelman

All we did, is kinda: $$\large\color{black}{ \displaystyle 3a\cdot 4x=12ax }$$

18. SolomonZelman

we multiplied the coefficients, but you can't really put the logarithms together more than "log(2)log(7)"

19. anonymous

yaa

20. SolomonZelman

21. anonymous

so we need to put the values of log 2 and log 7 to calculate the exact answer?

22. anonymous

as we cannt do anthing with them further

23. SolomonZelman

If you calculate these values it would be an approximation, and the "exact answer" is what we wrote before 98log(2)log(7)

24. anonymous

49log2*2log7=7^2log2*2log7=7log2^2*2log7=7*2log7=2 this is the solution by somebody on another site . pls see to it

25. SolomonZelman

No, it is not equivalent to 2. THAT IS WRONG....

26. anonymous

i didnt get the logic here

27. anonymous

yaa

28. SolomonZelman

no logic... it is wrong... if it had logic it would be much more frightening:O

29. anonymous

i have solved it , we have to take 2 as a base in frst case

30. SolomonZelman

I can't really follow what you did....

31. anonymous

i have dun blunder to this ques *O*

32. anonymous

hahaha

33. SolomonZelman

we can't take 2 as the base if it is a log.... you can't rewrite what's in the log, as the base... I mean you can't, but nothing correct is going to get out of it.

34. SolomonZelman

As again, the only thing here that I see valid is: $$\large\color{black}{ \displaystyle 49\log(2)~\cdot~2\log(7) =98\log(2)\log(7)}$$

35. anonymous

ok