## DarkBlueChocobo one year ago Help with logarithims

1. DarkBlueChocobo

2. DarkBlueChocobo

For the first one would you just fill in x for 0?

3. DarkBlueChocobo

to which we would get log(10)10+3

4. freckles

yes replace x with 0

5. freckles

$\log_a(a)=1 \text{ where } a \in (0,1) \cup (1,\infty)$

6. freckles

since a^1 is a

7. DarkBlueChocobo

I am confused are you saying log10(10)=1?

8. freckles

yes $10 \in (1,\infty) \\ \text{ so } 10 \in (0,1) \cup (1,\infty) \\ \text{ so yes } \log_{10}(10)=1$

9. DarkBlueChocobo

so like then we just add 3? so f(0)=4?

10. freckles

since 10^1=10

11. freckles

yep

12. DarkBlueChocobo

and for the second one would you just divide log(10)20/log(10)3

13. freckles

yes

14. DarkBlueChocobo

so log(3)20=x so you just seperate? and then they both have the common log(10)? im trying to udnerstand why

15. freckles

you can solve the equation by first taking log_10 of both sides

16. freckles

$\log_{10}(20)=\log_{10}(3^x) \\ \log_{10}(20)=x \log_{10}(3) \text{ by power rule }$

17. freckles

then divide both sides be log_10(3)

18. DarkBlueChocobo

Alrights Thank you for clearing that up. Egh last night of cramming for this test

19. freckles

$\frac{\log_{10}(20)}{\log_{10}(3)}=\frac{x \log_{10}(3)}{\log_{10}(3)} \\ \frac{\log_{10}(20)}{\log_{10}(3)}=x$

20. freckles

you could have chosen any log base as long as that base is a number between 0 and 1 exclusive or between 1 and infinity exclusive

21. freckles

maybe they want you to express in base 5 log well take log_5( ) of both side instead

22. freckles

just doing an example... $20=3^x \\ \log_5(20)=\log_5(3^x) \\ \log_5(20)=x \log_5(3) \\ \frac{\log_5(20)}{\log_5(3)}=x$

23. freckles

notice that answer could be expressed as: $x=\frac{\log_a(20)}{\log_a(3)} \text{ where } a \in (0,1) \cup (1,\infty)$

24. anonymous

can we say 3^x=20?

25. freckles

yes equality is symmetrical if a=b then b=a

26. anonymous

the can we say x=log3(20)?

27. freckles

yes my answer will also conclude that if you put a as 3 $x=\frac{\log_3(20)}{\log_3(3)}=\frac{\log_3(20)}{1}=\log_3(20)$

28. anonymous

can't we say x as a decimal?

29. freckles

you mean approximate x ?

30. anonymous

yep

31. anonymous

3^1=3 3^2=9 3^3=27 so x must be between 2 and three.... and so on?

32. anonymous

But if we are asked to find x for 4 decimal places can we find it? :(

33. freckles

calculator would probably be your friend

34. anonymous

can we find it without using it?