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anonymous

  • 5 years ago

If (log_3 X) (log_5 3) = 3, find x

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  1. anonymous
    • 5 years ago
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    Which one of these has an x?

  2. anonymous
    • 5 years ago
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    The other one is just a multiplicative constant. You can divide it over to the other side.

  3. anonymous
    • 5 years ago
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    ?? how? the bases are diff-- three and five...how can u just divide?

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

  5. anonymous
    • 5 years ago
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    =( i still don't get it....

  6. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    Yes, but first you have to get \(log_3x\) by itself.

  9. anonymous
    • 5 years ago
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    so it's log_3 x = 3c and then 3^3c=x

  10. anonymous
    • 5 years ago
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    Nope

  11. anonymous
    • 5 years ago
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    whaat???nooo

  12. anonymous
    • 5 years ago
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    C(log_3 x) = 3 So you have to divide by C, not multiply

  13. anonymous
    • 5 years ago
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    OH. lol...right.

  14. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    No.

  16. anonymous
    • 5 years ago
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    That part is trickier

  17. anonymous
    • 5 years ago
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    But lets see what you have with C for now.

  18. anonymous
    • 5 years ago
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    x=3^3/c

  19. anonymous
    • 5 years ago
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    Is that (3^3)/c or 3^(3/c) ?

  20. anonymous
    • 5 years ago
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    second one

  21. anonymous
    • 5 years ago
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    Correct. Ok, so now to find what C is. \[C = log_5 3 \iff 5^C = 3\]

  22. anonymous
    • 5 years ago
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    With me so far?

  23. anonymous
    • 5 years ago
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    yep

  24. anonymous
    • 5 years ago
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    Ok, so that means that if we take the ln of both sides we get \[C(ln\ 5) = (ln\ 3)\]

  25. anonymous
    • 5 years ago
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    polpak i need ur help please

  26. anonymous
    • 5 years ago
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    right?

  27. anonymous
    • 5 years ago
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    yep

  28. anonymous
    • 5 years ago
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    So that means \[C = \frac{(ln\ 3)}{(ln\ 5)}\]

  29. anonymous
    • 5 years ago
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    oh~~~

  30. anonymous
    • 5 years ago
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    Which means that \[log_53 = \frac{(ln\ 3)}{(ln\ 5)}\]

  31. anonymous
    • 5 years ago
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    oh, wow! thanks so much!!

  32. anonymous
    • 5 years ago
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    that makes so much more sense now.

  33. anonymous
    • 5 years ago
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    Glad I could help.

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