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pottersheep Group Title

Please help! Show that each statement is true (LOGS) [1/log base 5 of a] + [1/log base 3 of a] = [1/logbase 15a]

  • one year ago
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  1. pottersheep Group Title
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    sorry last one is log base 15 of a

    • one year ago
  2. satellite73 Group Title
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    lets add on the left

    • one year ago
  3. satellite73 Group Title
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    \[\frac{\log_3(a)+\log_5(a)}{\log_3(a)\times \log_5(a)}\]

    • one year ago
  4. satellite73 Group Title
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    now lets flip it and see if we can show it is the same as \(\log_{15}(a)\)

    • one year ago
  5. satellite73 Group Title
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    that is, can we show \[\frac{\log_3(a)\times \log_5(a)}{\log_3(a)+\log_5(a)}=\log_15(a)\] and i think we can do it using the change of base formula and writing everything in terms of \(\log_{15}(x)\)

    • one year ago
  6. satellite73 Group Title
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    do you know the change of base formula?

    • one year ago
  7. satellite73 Group Title
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    you get a nasty compound fraction on the left, i will see if i can write it \[\frac{\frac{\log_{15}(a)}{\log_{15}(5)}\times \frac{\log_{15}(a)}{\log_{15}(5)}}{\frac{\log_{15}(a)}{\log_{15}(5)}+\frac{\log_{15}(a)}{\log_{15}(3)}}\]

    • one year ago
  8. satellite73 Group Title
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    to clear the compound fraction, multiply top and bottom by \(\log_{15}(3)\log_{15}(5)\)

    • one year ago
  9. satellite73 Group Title
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    the numerator will be \[\log_{15}(a)\times \log_{15}(a)\] and the denominator will be \[\log_{15}(a)\log_{15}(5)+\log_{15}(a)+\log_{15}(5)\] factor as \[\log_{15}(a)\left(\log_{15}(5)+\log_{15}(3)\right)\] which gives \[\log_{15}(a)\times \log_{15}(3\times 5)=\log_{15}(a)\times 1\]

    • one year ago
  10. satellite73 Group Title
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    typo above, second line should be \[\log_{15}(a)\log_{15}(5)+\log_{15}(a)\log_{15}(5)\]

    • one year ago
  11. satellite73 Group Title
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    cancel and you get what you want

    • one year ago
  12. pottersheep Group Title
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    oooooo Thanks!

    • one year ago
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