## yeller33 3 years ago logx-log6=log15

1. sauravshakya

HINT: logx-log6 = log(x/6)

2. sauravshakya

Now, can u?

3. TheViper

so $$\frac{x}{6}=15$$ ?? @sauravshakya

4. sauravshakya

Yep... @TheViper

5. TheViper

Yeah I hoped :) THANX ;)

6. sauravshakya

WElcome

7. vishweshshrimali5

@yeller33 see, log a- log b = log(a/b) and log a + log b = log (ab) remember these 2 formula for ever

8. abayomi12

log(x) - log(6) = log(15) Add log(6) to each side: log(x) = log(15) + log(6) = log(15 times 6) x = 15 times 6 x = 90

9. vishweshshrimali5

Gud

10. sauravshakya

|dw:1345303238963:dw|And , right? @vishweshshrimali5

11. vishweshshrimali5

Yes @sauravshakya absolutely correct

12. yeller33

okay i understand thanks guys!

13. sauravshakya

Welcome

14. yeller33

how do you do it if it has a number at the end. = log7+logx=2.... the 2 throws me off.

15. Kaederfds

x is 90

16. sauravshakya

log7 + logx=2 log(7*x)=2

17. sauravshakya

Does it help?

18. vishweshshrimali5

and remember that $\large log_a a = 1$

19. yeller33

x=2//7?? ( 7x=2 )

20. vishweshshrimali5

Thus u can write 2 = 2* 1 = $$\large 2* log_{10} 10$$ also

21. sauravshakya

log7 + logx=2 log(7*x)=2 10^2 = 7*x 100/7=x

22. vishweshshrimali5

$\large b\space log _a x = log_a (b^x)$

23. vishweshshrimali5

That's how @sauravshakya got the third step

24. sauravshakya

got it? @yeller33

25. vishweshshrimali5

i can write 2 = $$\large \log_{10} (10^2)$$

26. sauravshakya

Yep @vishweshshrimali5

27. vishweshshrimali5

:)

28. sauravshakya

log7 + logx=2 log(7*x)=2 1/2 log(7x)=1 log{(7x)^1/2} = log10 (7x)^(1/2) = 10 7x = 100 x=100/7

29. sauravshakya

according to @vishweshshrimali5

30. yeller33

can you two see if i did this one correctly: logx+log8=1 log8*logx=1 8x=1 8x=10^1 10/8

31. yeller33

32. sauravshakya

dont think so

33. sauravshakya

34. sauravshakya

logx+log8= log8x

35. sauravshakya

and 1=log10

36. sauravshakya