## A community for students. Sign up today

Here's the question you clicked on:

## anonymous 5 years ago If (log_3 X) (log_5 3) = 3, find x

• This Question is Closed
1. anonymous

Which one of these has an x?

2. anonymous

The other one is just a multiplicative constant. You can divide it over to the other side.

3. anonymous

?? how? the bases are diff-- three and five...how can u just divide?

4. anonymous

Because the log base 5 of 3 is just a number. Like 4/5 or the square root of 15

5. anonymous

=( i still don't get it....

6. anonymous

Ok here. Lets say that $C=log_5 3$ Now your equation looks like: $C(log_3 x) = 3$ Can you solve it now?

7. anonymous

hmm..you know how log_b Y = x is equal the y=b^x in exponential form? how could you do that with this equation? isn't that how ur supposed to solve these things?

8. anonymous

Yes, but first you have to get $$log_3x$$ by itself.

9. anonymous

so it's log_3 x = 3c and then 3^3c=x

10. anonymous

Nope

11. anonymous

whaat???nooo

12. anonymous

C(log_3 x) = 3 So you have to divide by C, not multiply

13. anonymous

OH. lol...right.

14. anonymous

when i solve for (log_5 3)--to conv to exp form, does it still equal to three, which is the what the whole equation is equal to?

15. anonymous

No.

16. anonymous

That part is trickier

17. anonymous

But lets see what you have with C for now.

18. anonymous

x=3^3/c

19. anonymous

Is that (3^3)/c or 3^(3/c) ?

20. anonymous

second one

21. anonymous

Correct. Ok, so now to find what C is. $C = log_5 3 \iff 5^C = 3$

22. anonymous

With me so far?

23. anonymous

yep

24. anonymous

Ok, so that means that if we take the ln of both sides we get $C(ln\ 5) = (ln\ 3)$

25. anonymous

polpak i need ur help please

26. anonymous

right?

27. anonymous

yep

28. anonymous

So that means $C = \frac{(ln\ 3)}{(ln\ 5)}$

29. anonymous

oh~~~

30. anonymous

Which means that $log_53 = \frac{(ln\ 3)}{(ln\ 5)}$

31. anonymous

oh, wow! thanks so much!!

32. anonymous

that makes so much more sense now.

33. anonymous

Glad I could help.

#### Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy