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3psilon

  • 2 years ago

3^x = 5^y what is x/y ? Plz help . I'm terrible with Logs

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  1. CliffSedge
    • 2 years ago
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    Bah, logs with different bases . . . Gimme a sec to remember how to do this.

  2. CliffSedge
    • 2 years ago
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    Ok, got it. Take \(\large log_3\) of both sides.

  3. CliffSedge
    • 2 years ago
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    Then you can use the logarithm power rule.

  4. 3psilon
    • 2 years ago
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    Do you mind working please? @CliffSedge . I'm a little rusty on the rules

  5. CliffSedge
    • 2 years ago
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    Oh, I suppose so. It's pretty easy.. Given \(\large 3^x = 5^y\) \(\large log_3\) both sides \(\large log_3(3^x) = log_3(5^y)\) \(\large x = 3^x = log_3(5^y)\) Understand so far?

  6. CliffSedge
    • 2 years ago
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    *oops, never mind that = "3x" in the last line. I thought I deleted that.

  7. 3psilon
    • 2 years ago
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    I'm sorta catching on

  8. CliffSedge
    • 2 years ago
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    Should be \(\large x=log_3(5^y)\)

  9. CliffSedge
    • 2 years ago
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    Do you see that \(\large log_3(3^x)=x\) ?

  10. CliffSedge
    • 2 years ago
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    The logarithm function tells you what the exponent is on that base.

  11. CliffSedge
    • 2 years ago
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    So, since a logarithm is an exponent, and if you have a power-to-a-power, you multiply exponents, \(\large log_3(5^y)=y log_3(5)\) . I think you can get the rest from here.

  12. zepdrix
    • 2 years ago
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    |dw:1351900584391:dw| Still having some trouble pimp or you figured this one out already. Figured I'd throw my 2 cents just in case :D Sorry that it's a little bit messy, hopefully you can follow it. The asterisks are log rules.

  13. CliffSedge
    • 2 years ago
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    I don't think the change of base is necessary, but it is good to know how to do such a thing.

  14. 3psilon
    • 2 years ago
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    I found this way to do it|dw:1351900905598:dw|

  15. zepdrix
    • 2 years ago
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    If we wants to be able to plug it into a calculator it is necessary :) Depends what type of answer his teacher wants I suppose! :D

  16. zepdrix
    • 2 years ago
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    wow that's tough to read :(

  17. CliffSedge
    • 2 years ago
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    That is an equivalent way, sure. It uses the same log rules but in a different order.

  18. 3psilon
    • 2 years ago
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    Sorry I just put them both to the 1/y

  19. zepdrix
    • 2 years ago
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    took the y'th root? Oh that's clever ^^

  20. CliffSedge
    • 2 years ago
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    Makes sense since it's looking for a ratio of exponents, but also takes a leap of imagination that might be hard to see. The straight-forward way is easier to remember and just as efficient.

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