## anonymous 4 years ago (logx)^(logx)=? pfft didn't do log for quite a while :P

1. Mertsj

$\log a ^{n}=nloga$

2. anonymous

not logx^(logx)

3. myininaya

$\text{ Let } y=(\log(x))^{\log(x)}$ Now do log( ) of both sides. $\log(y)=\log[(\log(x))^{\log(x)}]$ Using properties of log we can rewrite as $\log(y)=\log(x)\cdot \log(\log(x))$ Again we can use another property to rewrite this as Now I will write this as an exponential equation where the base is 10 assuming the base is 10 here. $10^{\log(y)}=10^{\log(x)\cdot \log(\log(x))}$ $y=10^{\log(x) \cdot \log(\log(x))}$ => $(\log(x))^{\log(x)}=10^{\log(x) \cdot \log(\log(x))}$ Both still pretty ugly, but I honestly don't know what you want to show here.

4. anonymous

oh no, i have to proof its = x

5. myininaya

what is the base?

6. myininaya

i assumed it was 10

7. anonymous

10 i think question didnt write

8. anonymous

if it was another base im'm sure it wouldn't affect the answer

9. myininaya

The base would just be some other number if not 10

10. myininaya

$(\log(x))^{\log(x)}=a^{\log(x) \cdot \log(\log(x))}$ if the base was a

11. anonymous

Solve the equation (logx)^logx=x

12. anonymous

thats how the question says

13. myininaya

Oh we are solving an equation

14. anonymous

log(logx)^logx=logx (logx)log(logx)=logx I cant cancel log on both sides right?

15. bahrom7893

yea u can

16. myininaya

$\log(x)(\log(\log(x))-1)=0$

17. myininaya

subtract log(x) on both sides

18. myininaya

both terms have log(x) in common

19. myininaya

you factor

20. anonymous

log(logx)=1 log(logx)=log10 logx=10 x=100 right?

21. myininaya

what about x=10 for log(x)=0 since log(10)=1

22. anonymous

oh wait log 100 isnt 10

23. anonymous

x= 10^10!

24. myininaya

$x=10^{10}$

25. myininaya

you have two solutions

26. anonymous

whats the other solution?

27. anonymous

i took out logx so logx=0 right

28. anonymous

or logx=1

29. myininaya

$(\log(x))^{\log(x)}=x$ D log( ) on both sides $\log(x) \cdot \log(\log(x))=\log(x)$ Now subtract log(x) on both sides $\log(x) \cdot \log(\log(x))-\log(x)=0$ Now factor the expression on the left side $\log(x)(\log(\log(x))-1)=0$ Now set both factors =0 So we have $\log(x)=0 \text{ or } \log(\log(x))-1=0$ So we should solve both equations $x=10 \text{ or } \log(\log(x))=1$ $x=10 \text{ or } 10^{\log(\log(x))}=10^1$ $x=10 \text{ or } \log(x)=10$ $x=10 \text{ or } 10^{\log(x)}=10^{10}$ $x=10 \text{ or } x=10^{10}$

30. myininaya

But if we check both solutions....

31. anonymous

I divided the log away instead of subtracting it,now i get where i was wrong, thanks a lot that was very kind of you!

32. anonymous

cancel the 10 right?

33. Zarkon

I think you ment to say x=1 though after checking one sees that is not a solution

34. bahrom7893

Travis you could have divided, you just made a mistake here: log(logx)=1 log(logx)=log10 logx=10 then x !=100

35. bahrom7893

10^log(x) = 10^10 x = 10^10

36. myininaya

oh yeah zarkon for some rason i solve log(x)=1

37. myininaya

reason*

38. anonymous

Yea I was too busy manipulating the log I didn't convert the log properly

39. myininaya

log(x)=0 x=1 for sure! :)

40. anonymous

yea so x =1 and x= 10^10

41. Zarkon

x=1 is only a solution if you define $0^0=1$

42. myininaya

right which i don't define it that way

43. anonymous

but if you sub x = 10^10 it works too

44. myininaya

i would say that is the only solution

45. myininaya

because after checking x=1 we get 0^0 like zarkon said

46. anonymous

So reject that solution right